What possible error can occur by using (a) a single large piece of ice or (b) very fine shavings of ice in an experiment for the latent heat of fusion of ice?

1. What possible error can occur by using (a) a single large piece of ice or (b) very fine shavings of ice in an experiment for the latent heat of fusion of ice?

Answer:

Energy is required to change water from a solid to a liquid, i.e. to melt ice. In this experiment you will try to measure the latent heat of fusion of ice (LHice), the energy needed (per gram) to melt ice. The needed energy will come from a cup of warm water. The amount of water and its temperature will be measured before adding some ice and then again after the ice has been melted. These data will be used in an energy balance equation to determine LHice.

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2. What possible error can occur by using (a) a single large piece of ice or (b) very fine shavings of ice in an experiment for the latent heat of fusion of ice?

Errors are normally classified in three categories: systematic errors, random errors, and blunders. Systematic errors are due to identified causes and can, in principle, be eliminated. Errors of this type result in measured values that are consistently too high or consistently too low.

3. what is the difference latent heat of fusion and latent heat of vaporization?​

Answer:

Latent heat of fusion is the amount of heat required to change from solid to liquid to solid.Likewise,Latent heat of vaporization is the amount of heat needed to change liquid to vapor or released when vapor condenses back to liquid

4. A Styrofoam box used to keep drinks cold at a picnic has a total wall area (including the lid) of 0.80 m2and wallthickness 3.0 cm. It is filled with ice, water, and cans of Coca-Cola at 0°C. (a) What is the rate of heat flow intothe box if the temperature of the outside wall is 30°C? (b) How much ice melts in one day? (The latent heat offusion of ice is 334,000 J/kg.)

Answer:

12 watts is 12 joules per second, and over 24 hours, that is

12 J/s x 3600s/hour x 24 hr/day = 1.037e6 joules

heat of fusion of ice is 334 kJ/kg

1.037e6 joules = 334000 J/kg x M

M = 3.1 kg

Explanation:

thats the answer?

5. How much heat is absorbed by 100g of ice at -10⁰C to become water at 20⁰C? • Specific heat: Water (Ice) = 2,090 J/Kg•⁰C, Water (Liquid) = 4,186 J/Kg•⁰C • Latent heat of fusion (Water) = 3.33 x10^5 J/Kg _____ J/Joules

How much heat is absorbed by 100g of ice at -10⁰C to become water at 20⁰C?

• Specific heat: Water (Ice) = 2,090 J/Kg•⁰C, Water (Liquid) = 4,186 J/Kg•⁰C

• Latent heat of fusion (Water) = 3.33 x10^5 J/Kg

The amount of heat absorbed by 100 grams of ice at -10⁰C into water at 20⁰C is 43,763 J/Joules

To solve this problem, you can use the heat formula.

Heat is a form of energy that can move from an object with a higher temperature to an object with a lower temperature if the two are brought together.

The Heat Formula

1. Heat Transfer Formula

Q = m x c x ΔT

2. Specific Heat Formula

c = Q/(m x ΔT)

3. Heat of Melting and Vapor Formula

Heat of fusion

Q = m x L

Steam heat

Q = m x U

Information:

c = specific heat of substance (J/kg⁰C)

Q = the amount of heat released or received by an object (Joules)

m = mass of the object receiving or releasing heat (kg)

T = change in temperature (⁰C)

Explanation with steps:

Given:

mass = 100 g ⇒ 0.1 KgSpecific heat: Water (Ice) = 2,090 J/Kg•⁰C, Water (Liquid) = 4,186 J/Kg•⁰CLatent heat of fusion (Water) = 3.33 x10^5 J/Kg

Question:

How much heat is absorbed by 100g of ice at -10⁰C to become water at 20⁰C?

Solution:

Step 1

Find the amount of heat when the ice melts at -10⁰C to become at 0⁰C

Q₁ = m x c x ΔT

Q₁ = 0.1 Kg x 2,090 J/Kg•⁰C x (0 - (-10)⁰C

Q₁ = 2,090 J

Step 2

Find the amount of heat when the ice melts

Q₂ = m x L

Q₂ = 0.1 Kg x 3.33 x 10⁵ J/Kg

Q₂ = 3.33 X 10⁴ J

Q₂ = 33,300 J

Step 3

Find the amount of heat when the ice melts at 0⁰C to become at 20⁰C

Q₃ = m x c x ΔT

Q₃ = 0.1 Kg x 4,186 J/Kg•⁰C x (20 - 0)⁰C

Q₃ =  0.1 Kg x 4,186 J/Kg•⁰C x 20⁰C

Q₃ = 8,372 J

Step 4

Find the amount of heat needed.

Q total = Q₁ + Q₂ + Q₃

Q total = 2,090 J + 33,300 J + 8,372 J

Q total = 43,763 J

The amount of heat absorbed by 100 grams of ice at -10⁰C into water at 20⁰C is 43,763 J/Joules

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6. What's the latent heat of fusion of water?

Answer:

Similarly, while ice melts, it remains at 0 °C (32 °F), and the liquid water that is formed with the latent heat of fusion is also at 0 °C. The heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram, and the heat of vaporization at 100 °C is about 2,230 joules (533 calories) per gram.

7. 6. How many kilograms of ice at 0°C can be melted to water at 0 °C with 10 kilo calories of heat? Heat of fusion of ice is 80 cal/gm and 1 kcal = 1000 cal​

Answer:

To solve this problem, we need to use the heat equation:

Q = m × L

Where Q is the amount of heat required, m is the mass of the substance, and L is the specific heat of the substance.

In this case, we need to find the mass of ice that can be melted to water with 10 kilo calories of heat. We know that the heat of fusion of ice is 80 cal/gm, which means that it takes 80 calories of heat to melt one gram of ice at 0°C.

First, we need to convert 10 kilo calories to calories:

10 kilo calories = 10,000 calories

Now we can use the heat equation to find the mass of ice:

10,000 = m × 80

m = 10,000 / 80

m = 125 grams

So, 125 grams of ice at 0°C can be melted to water at 0°C with 10 kilo calories of heat.

To convert this to kilograms, we need to divide by 1000:

125 grams = 0.125 kilograms

Therefore, 0.125 kilograms of ice at 0°C can be melted to water at 0°C with 10 kilo calories of heat.

8. 1. 5 questions regarding specific heat with answer and complete solution.2. 5 questions regarding latent heat of fusion with answer and complete solution. 3. 5 questions regarding latent heat of vaporization with answer and complete solutionpa help plssss​

Answer:

Questions regarding specific heat:

Question 1: A 500 g aluminum block is heated from 25°C to 75°C. If the specific heat of aluminum is 0.91 J/g°C, how much heat energy is required?

Answer:

Given:

Mass of aluminum block = 500 g

Initial temperature of aluminum block = 25°C

Final temperature of aluminum block = 75°C

Specific heat of aluminum = 0.91 J/g°C

To find:

Heat energy required

Solution:

Heat energy required = mass x specific heat x change in temperature

Change in temperature = final temperature - initial temperature

= 75°C - 25°C

= 50°C

Heat energy required = 500 g x 0.91 J/g°C x 50°C

= 22,750 J

Therefore, the heat energy required to heat the 500 g aluminum block from 25°C to 75°C is 22,750 J.

Question 2: A 2.5 kg block of copper is heated from 20°C to 80°C. If the specific heat of copper is 0.385 J/g°C, how much heat energy is required?

Answer:

Given:

Mass of copper block = 2.5 kg

Initial temperature of copper block = 20°C

Final temperature of copper block = 80°C

Specific heat of copper = 0.385 J/g°C

To find:

Heat energy required

Solution:

Convert mass to grams:

Mass of copper block = 2.5 kg x 1000 g/kg

= 2500 g

Change in temperature = final temperature - initial temperature

= 80°C - 20°C

= 60°C

Heat energy required = mass x specific heat x change in temperature

= 2500 g x 0.385 J/g°C x 60°C

= 57,750 J

Therefore, the heat energy required to heat the 2.5 kg copper block from 20°C to 80°C is 57,750 J.

Question 3: How much heat energy is required to raise the temperature of 1 liter of water from 20°C to 90°C? The specific heat of water is 4.18 J/g°C.

Answer:

Given:

Volume of water = 1 L

Initial temperature of water = 20°C

Final temperature of water = 90°C

Specific heat of water = 4.18 J/g°C

To find:

Heat energy required

Solution:

The mass of water can be calculated using its density:

Density of water = 1 g/mL

Mass of water = volume x density

= 1 L x 1000 g/L

= 1000 g

Change in temperature = final temperature - initial temperature

= 90°C - 20°C

= 70°C

Heat energy required = mass x specific heat x change in temperature

= 1000 g x 4.18 J/g°C x 70°C

= 293,200 J

Therefore, the heat energy required to raise the temperature of 1 liter of water from 20°C to 90°C is 293,200 J.

Question 4: How much heat energy is required to melt 500 g of ice at 0°C? The latent heat of fusion of ice is 334 J/g.

Answer:

Given:

Mass of ice = 500 g

Latent heat of fusion of ice = 334 J/g

To find:

Heat energy required

Solution:

Heat energy required to melt ice = mass x latent heat of fusion

= 500 g x 334 J/g

= 167,

9. examples of latent heat of sublimation​

Answer:

The latent heat of sublimation at a particular temperature is the amount of heat required to convert a unit mass of solid into gas. For example, when ice sublimates into vapor, the amount of heat required at 0°C is estimated equal to 2,838 kJ/kg, which is the latent heat of sublimation of ice at 0°C.

10. Recalculate the latent heat of fusion, assuming that 5.0 grams of water were transferred to the calorimeter on the surface of the ice. Is this the expected result? Why?

Explanation:

Find the latent heat of fusion, Lf, according to Lf = q ÷ m by dividing the heat, q, absorbed by the ice, as determined in step 3, by the mass of ice, m, determined in step 4. In this case, Lf = q / m = 2293 J ÷ 7.0 g = 328 J/g. Compare your experimental result to the accepted value of 333.5 J/g

11. How much energy is needed to melt 15kg of ice at 0oC to water at the same temperature? (Specific latent heat of fusion of water = 336000 j/kg)

Answer:

To melt 15 kg of ice at 0°C to water at the same temperature, we need to use the following formula:

Q = mL

Where:

Q = amount of energy required (in joules)

m = mass of ice (in kg)

L = specific latent heat of fusion of water (in J/kg)

Substituting the given values:

Q = 15 kg x 336000 J/kg

Q = 5,040,000 J

Therefore, 5,040,000 J of energy is needed to melt 15 kg of ice at 0°C to water at the same temperature.

12. Solve using your paper. Question An 10 kg ice was turned to water vapor. how much heat is needed to do that given that specific heat of water is 4, 186 J/kg*C , Latent Heat of Fusion is 304553.36 J/kg and latent heat of vaporization is 2260000 J/kg. PLEASE SHOW YOUR SOLUTION

Answer:

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Answer:

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Explanation:

salamat po ha

13. Define the term latent heat of fusion

Definition of latent heat of fusion. The amount of heat required to change 1 g of a substance at the temperature of its melting point from the solid to the liquid state without changing temperature.

the heat absorbed by a unit mass of a givensolid at its melting point that completelyconverts the solid to a liquid at the sametemperature: equal to the heat ofsolidification.

14. What is latent heat?

Answer:

Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process — usually a first-order phase transition. Latent heat can be understood as energy in hidden form which is supplied or extracted to change the state of a substance without changing its temperature.

15. 10. Why does the bottom of a lake not freeze in severe winter even when the surface is all frozen? explainA. The water has a large specific heatB. The conductivity of ice is lowC. The water has large latent heat of fusionD. The temperature of the earth at the bottom of the lake is high​

Answer:

the pressure increases. ... Also as ice is less dense as compared to water that is why ice always floats on water and doesn't settle down at the bottom or that is the reason why the lake doesn't start to freeze from the bottom.

Explanation:

16. An 10 kg ice was turned to water vapor. how much heat is needed to do that given that specific heat of water is 4, 186 J/kg*C , Latent Heat of Fusion is 304553.36 J/kg and latent heat of vaporization is 2260000 J/kg.

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17. how to solve for latent heat

In physics, latent heat is the heat per kilogram that you have to add or remove to make an object change its state; in other words, latent heat is the heat needed to make a phase change happen. Its units are joules per kilogram (J/kg) in the MKS (meter-kilogram-second) system.

Physicists recognize three types of latent heat, corresponding to the changes of phase between solid, liquid, and gas:

The latent heat of fusion, Lf. This is the heat per kilogram needed to make the change between the solid and liquid phases, as when water turns to ice or ice turns to water.

The latent heat of vaporization, Lv. This is the heat per kilogram needed to make the change between the liquid and gas phases, as when water boils or when steam condenses into water.

The latent heat of sublimation, Ls. This is the heat per kilogram needed to make the change between the solid and gas phases, as when dry ice evaporates.

image0.png

Here’s the formula for heat transfer during phase changes, where

image1.png

m is the mass, and L is the latent heat:

image2.png

Here, L takes the place of the

image3.png

and c (specific heat) terms in the temperature-change formula.

Suppose you’re in a restaurant with a glass of 100.0 grams of water at room temperature, 25 degrees Celsius, but you’d prefer ice water at 0 degrees Celsius. How much ice would you need? You can find the answer using the heat formulas for both change in temperature and phase change.

You get out your clipboard, reasoning that the heat absorbed by the melting ice must equal the heat lost by the water you want to cool. Here’s the heat lost by the water you’re cooling:

image4.png

T is the final temperature, and T0 is the initial temperature.

Plugging in the numbers tells you how much heat the water needs to lose:

image5.png

So how much ice would that amount of heat melt? That is, how much ice at 0 degrees Celsius would you need to add to cool the water to 0 degrees Celsius? That would be the following amount, where Lf is the latent heat of fusion for ice:

image6.png

You know this has to be equal to the heat lost by the water, so you can set this equal and opposite to

image7.png

In other words,

image8.png

“Pardon me,” you say to the waiter. “Please bring me exactly 31.0 grams of ice at precisely 0 degrees Celsius.”

18. If 2500kJ of energy is just enough to melt 3.0kg of a substance, what is the substance’s latent heat of fusion?

Answer:

It's the quantity of heat needed to raise the temperature of a substance by a given amount. The amount of heat produced is determined by the substance's characteristics. This means that the amount of heat produced by various substances will differ plus Latent heat energy (symbol '(l)') is the name given to this type of heat energy. The specific latent heat of a substance is the amount of thermal energy necessary to change the state of 1 kg of the substance at its melting point.

19. Judy places 0.150 kg of boiling water in a thermos bottle. How many kilograms of ice at –12.0 °C must Judy add to the thermos so that the equilibrium temperature of the water is 75.0 °C? specific heat of ice = 2090 J/kg·°C specific heat of water = 4186 J/kg·°C Latent heat of fusion of water = 3.34x105 J/kg

Answer:I assume by "boiling water" you mean water at 100ºC. It can't boil without an input of energy.

specific heat of water is 4.186 kJ/kgC

specific heat of ice is 2.06 kJ/kgC

specific heat of steam is 2.1 kJ/kgK

heat of fusion of ice is 334 kJ/kg

heat of vaporization of water is 2256 kJ/kg

Energy lost by the ice in warming up to 75º must equal that lost by the water in cooling from 100º to 75º

E1 =  2.06 kJ/kgC × M × 12º

E2 = 334 kJ/kg × M

E3 = 4.186 kJ/kgC × M × 75º

E4 = 4.186 kJ/kgC × 0.15kg × (100–75)

E1+E2+E3 = E4

solve for M

Explanation:

20. Suppose that 5.0 grams of water is transferred to the calorimeter on the surface of the ice and assuming that all data are correct, would the calculated value for the latent heat of fusion be too high or too low

ANSWER:

So you've transferred 5.0 g of water at O deg C along with the ice at O deg C. What measurements will this directly affect? And how will they be affected? And what does that do to the quantities in the steps of the calculation and how is the result affected?

Let's look at the calculation of the latent heat of fusion:

You plug in all of the requisite values and solve for the unknown, which would be your experimental result. But if 5.0 g of cold water came along with the ice, this water as far as the measurements are concerned, will look like it is ice, not liquid, because you calculate the mass of ice by subtracting the mass of the calorimeter and warm water from the final mass of the calorimeter and its contents. That means the mass of ice is really 5.0 g less than what you get from this subtraction. To get the result for the latent heat of fusion corrected for this error, you must plug in a mass for ice that is 5.0 g less than if the error had not happened. But that is not all. This 5.0 g of cold water does warm up to the final temperature in the calorimeter. So we have to add a term to our guiding equation:

If you now calculate the heat of fusion of ice with this new equation, you'll be able to see if your original answer was too small or too big. But even more elegant is to logically deduce what happens just be looking at the setup for the original calculation:

Can you see that the mass of ice in the denominator is too big because you're assuming that the 5.0 g of water was actually ice? What would too large of a number in the denominator do to the result? And can you see that the mass of the cold water is actually ok because that really was the mass of the cold water in the calorimeter? See if your logical answer agrees with what you saw with your calculation check.

21. what is latent heat of vaporization?

Latent heat of vaporization is a physical property of a substance. It is defined as the heat required to change one mole of liquid at its boiling point under standard atmospheric pressure. It is expressed as kg/mol or kJ/kg. When a material in liquid state is given energy, it changes its phase from liquid to vapor; the energy absorbed in this process is called heat of vaporization. The heat of vaporization of water is about 2,260 kJ/kg, which is equal to 40.8 kJ/mol.

22. 10. Why does the bottom of a lake not freeze in severe winter even when the surface is all frozen? A. The water has a large specific heat B. The conductivity of ice is low C. The water has large latent heat of fusion D. The temperature of the earth at the bottom of the lake is high

Answer:

D. The temperature of thw earth at the bottom of the lake is high

Explanation:

sana makatulong

23. to a ___________ . This heat energy is called the latent heat of fusion.

Answer:

sorry diko po alam answer btw ty sa points

24. Suppose that 5.0 grams of water is transferred to the calorimeter on the surface of the ice and assuming that all data are correct, would the calculated value for the latent heat of fusion be too high or too low? Give reasons for your answer.

Explanation:

So you’ve transferred 5.0 g of water at 0 deg C along with the ice at 0 deg C. What measurements will this directly affect? And how will they be affected? And what does that do to the quantities in the steps of the calculation and how is the result affected?

Let’s look at the calculation of the latent heat of fusion:

You plug in all of the requisite values and solve for the unknown, which would be your experimental result. But if 5.0 g of cold water came along with the ice, this water, as far as the measurements are concerned, will look like it is ice, not liquid, because you calculate the mass of ice by subtracting the mass of the calorimeter and warm water from the final mass of the calorimeter and its contents. That means the mass of ice is really 5.0 g less than what you get from this subtraction. To get the result for the latent heat of fusion corrected for this error, you must plug in a mass for ice that is 5.0 g less than if the error had not happened. But that is not all. This 5.0 g of cold water does warm up to the final temperature in the calorimeter. So we have to add a term to our guiding equation:

If you now calculate the heat of fusion of ice with this new equation, you’ll be able to see if your original answer was too small or too big. But even more elegant is to logically deduce what happens just be looking at the setup for the original calculation:

Can you see that the mass of ice in the denominator is too big because you’re assuming that the 5.0 g of water was actually ice? What would too large of a number in the denominator do to the result? And can you see that the mass of the cold water is actually ok because that really was the mass of the cold water in the calorimeter? See if your logical answer agrees with what you saw with your calculation check.

25. How much heat energy (cal) is required to vaporize a 1.0-g ice cube at 0°c? the heat of fusion of ice is 80. cal/g. the heat of vaporization of water is 540 cal/g, and c = 1.0 cal/g°c.

Answer:

How much heat energy (cal) is required to vaporize a 1.0-g ice cube at 0°c? the heat of fusion of ice is 80. cal/g. the heat of vaporization of water is 540 cal/g, and c = 1.0 cal/g°c.

Explanation:

I hope this helps you

26. calculate the entropy changes, ∆S melting of ice at 0°C (heat of fusion of ice is 6.02 KJ/mol​

Answer:

Kween Mhaldita Juliana Grace Elaine Quices Quisquino Celestra

27. 5. the amount of thermal energy needed to raise the temperature of 1 kg of material by 1 degree Celsius is called the____. A. heat capacity B. specific heat capacity. C. latent heat of fusion. D. latent heat of vaporization​

Answer:

b.specific heat capacity

28. .What happens when liquid substance turns into solid?a. It loses energy equal to its latent heat of fusion.b. It absorbs energy equal to its latent heat of fusion.c. It losses energy equal to its latent heat of vaporization.d. It absorbs energy equal to its latent heat of vaporization.​

Answer:

A

Explanation:

trivia, the energy released upon the freezing(to change liquid substance into solid, freezing method can be used)it. Enthalphy fusion is a latent heat, it's exactly the same as the energy required to melt the same amount of solid. We are talking about liquid substance that turns into solid meaning to say, the liquid substance need to freeze to transform itself to solid. The latent heat are released because we are talking about freezing and not melting.

29. what is the difference between latent heat of fusion and latent heat of vaporization​

Answer:

The latent heat of fusion is the heat required to change the solid from solid to liquid state without any change in temperature. ... While Latent heat of vaporization is the heat required to change the liquid from liquid to vapor state without any change in temperature.

30. is latent heat related to specific heat?Why?​

Answer:

Specific heat: = energy required to change a unit mass of a material by 1°C. ... Latent heat = energy required to change the state (gas, liquid, solid) of a unit mass of material. Units: energy per unit mass.