How many calories does it take to heat 10 grams of ice to 3 degrees C?

Question

Can you explain it to me? I have never learned this and need help with explaining how to do this.

Answer 1

In order to answer this question, we need to know the starting temperature of the ice, which is not provided. We will assume it is 0ºC.

The first thing that must happen is that the 10 g of ice must MELT @ 0ºC. This is called a phase change because the ice is changing phases from solid to liquid. When this happens, all of the heat (calories or joules) that is added goes into melting the ice, but not changing the temperature. To calculate how many calories this takes, we need to know the ∆H_fusion of ice. This values is provided in tables that you can look up. I find it to be 79.83 cal / g.

So, to melt 10 g of ice it will take 79.83 cal / g x 10 g = 798.3 cal. The equation that is generally used for this calculation is q = m∆H_fus where q = heat (cal, or joules) and m = mass.

The next step is to now raise the temperature of the liquid water from 0ºC to 3ºC. The equation that is used here is q = mC∆T where q = heat (cal or joules), m = mass, C = specific heat of liquid water and ∆T = change in temperature. The C is usually given or you can look it up. It is ! = 1 cal / gº. Solving for q, we have...

q = (10 g)(1 cal / gº)(3º)

q = 30 cal

Adding up the heat from the 2 steps we can arrive at the total amount of heat (in calories) requred:

798.3 cal + 30 cal = 828 cal (not corrected for sig. figs.)