Specific heat capacity

Question

Find the quantity of heat needed to change a 1.0-gg ice cube at absolute zero (-273∘C∘C) to 1.0 gg of boiling water. Express your answer in calories.

Answer 1

Hello,

Assuming you would like to keep the specific heat capacity of water constant in liquid state. The Cp is 1 cal/gr-C. This means that it takes 1 calorie to raise the temperature of 1 gram of water by 1 degree C.

I assume that boiling will be set at 100 C.

Therefore,

Q = mCp(Tf-Ti) = (1 gr)(1 cal/gr-C)(100C-(-273C)) = 373 cal.

Now, If you want to consider the variation of specific heat as a function of temperature, the Cp of ice is 0.5 cal/gr-C

Then, we can break it into 3 parts:

Q = m[Cp_ice(T_melt-T_abs0) + Cp_liquid(T_boil-T_melt)] = (1 gr)[(0.5 cal/gr-C)(0-(-273)) + (1 cal/gr-C)(100-0)] = (1 gr)[136.5 cal/gr + 100 cal/gr] = 236.5 cal.

Hope this helps!

Dr. Chuy