Problem

The latent heat of fusion (solid become liquid) for ice is 79.8 cal/gram. How much ice can we melt with 1000 calories of heat?

Problem
Which would require more heat, melting 500 g of 0 C ice or turning 500 g of 100 C water into steam?
Problem
A 500 g sample of an unknown material requires 750 calories of heat to raise its temperature by 50 C. Use Equation (1) and Table 6.1 to identify the material.

Heat=Lf*mass

solve for mass.

In the second, compare the heats of fusion/vaporization. One is about six times as great as the other.

You will have to do it yourself. Find the heat capacity, and look in the table.

The latent heat of fusion (solid become liquid) for ice is 79.8 cal/gram. How much ice can we melt with 1000 calories of heat

To solve these problems, we need to understand the concepts of latent heat and specific heat.

1. Latent Heat of Fusion: This is the amount of heat energy required to change a substance from solid to liquid without changing its temperature. For ice, the latent heat of fusion is 79.8 cal/gram.

2. Specific Heat: This is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius. It is usually measured in calories per gram Celsius (cal/g°C).

Now, let's solve each problem step by step:

Problem 1: How much ice can we melt with 1000 calories of heat?

To solve this, we need to use the latent heat of fusion formula:

Amount of ice melted = Heat energy supplied / Latent heat of fusion

Substituting the values given in the problem:

Amount of ice melted = 1000 cal / 79.8 cal/gram

Calculating this, we get:

Amount of ice melted = 12.53 grams (rounded to two decimal places)

Therefore, we can melt approximately 12.53 grams of ice with 1000 calories of heat.

Problem 2: Which would require more heat, melting 500 g of 0°C ice or turning 500 g of 100°C water into steam?

To compare the amount of heat required in each process, we need to calculate the heat energy using the following formulas:

Heat energy for melting ice = Mass of ice × Latent heat of fusion
Heat energy for turning water into steam = Mass of water × Latent heat of vaporization

Substituting the values given in the problem:

Heat energy for melting ice = 500 g × 79.8 cal/g

Calculating this, we get:

Heat energy for melting ice = 39,900 calories

Heat energy for turning water into steam = 500 g × 540 cal/g

Calculating this, we get:

Heat energy for turning water into steam = 270,000 calories

Comparing the two heat energies, we can see that turning 500 g of 100°C water into steam requires more heat energy compared to melting 500 g of 0°C ice.

Problem 3: A 500 g sample of an unknown material requires 750 calories of heat to raise its temperature by 50°C. Use Equation (1) and Table 6.1 to identify the material.

To identify the material, we need to use the specific heat equation:

Heat energy = Mass × Specific heat × Change in temperature

Substituting the given values in the problem:

750 cal = 500 g × Specific heat × 50°C

Simplifying the equation:

Specific heat = (750 cal) / (500 g × 50°C)

Calculating this, we get:

Specific heat = 0.03 cal/g°C

Now, we can refer to "Table 6.1" to identify the material that has a specific heat of 0.03 cal/g°C. The table will provide a list of substances and their corresponding specific heat values.