need help please?

Question

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?
(a)Read the introduction carefully. Identify the key pieces of information required to solve this problem.
(b)Identify the extraneous information in the background reading
c) Explain your strategy for solving this problem.
(d)Solve the problem. How much ice can you melt with 1000 calories of heat?
Hints:

The formula is:
The amount of heat = (mass of the material) x (latent heat of fusion):
Q = mLf
where Q is the amount of heat in calories;
m is the mass of the material in grams;
Lf is the latent heat of fusion in calories/gram.

In this problem, you are provided the amount of heat, Q, and the latent heat of fusion, Lf. You will have to rearrange this equation and solve it for mass, m.

Which would require more heat, melting 500 g of 0 C ice or turning 500 g of 100 C water into steam?
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.The formula used to calculate the amount of heat required to heat the material without phase change is:

Amount of heat required = (specific heat capacity) x (mass of the material) x (change in temperature)

or

Q = C * m * ∆T

where C is the specific heat capacity of the material in cal/g 0C;
m is the mass of the material in grams;
∆T is the temperature change, 0C.

Answer 1

For the first question: How much ice can be melted with 1000 calories of heat, simply use the formula that you have, solving for the mass of ice. You are given the heat of fusion, which is the amount of heat required to melt 1 gram of ice. So,

1000 cal = m (79.8 cal/g)
m= 1000 cal/(79.8 cal/g)= 12.5 grams of ice.

In order to determine which requires more heat, melting 500 g of ice or vaporizing 500 grams of water, you'll need to have the heat of vaporization of water, which I do not see in the information given. I'm guessing that that value is available to you, though. In each case, you'll use a comparable equation, and in each case you will solve for "q".

In the third question, you can calculate the specific heat very easily, and then compare that value with the values in the table you are given.

Simply use the equation Q = C * m * ∆T and solve for C.