# I try to solve a thermodynamics problem on heat transfer

• Salmone
In summary, the thermodynamic container must have a heat capacity of 95740 Joules/kilogram to absorb the amount of heat given up by the water when the temperature is raised from -4 degrees to 40 degrees Celsius.

#### Salmone

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Homework Statement
Find the thermal capacity of a container with ice and water
Relevant Equations
##Q=C \Delta T##
I have some doubts about a thermodynamics exercise I did-I'll write down the text and show how I tried to solve it to see if I reasoned it out right.

In a container of thermal capacity ##C## containing ##1.0 kg## of ice at a temperature of ##-4.0 °C##, a mass of ##3.0 kg## of water is poured into it at a temperature of ##40.0 °C##. There is no heat exchange with the surroundings. The specific heats of water and ice are ##4186 J/kgK## and ##2090 J/kgK##, respectively, while the latent heat of fusion of ice is ##3.33 × 105 J/kg##.

The exercise asks to calculate the thermal capacity ##C## that the container must have if we want that at the equilibrium temperature just ##\frac{1}{3}## of the mass of the ice is melted.

I reasoned like this:

at the equilibrium temperature if the ice is not completely melted, it means the equilibrium temperature is 0 degrees. This means that the 3 kilograms of water will give up a total amount of heat equal to ##Q=cm \Delta T=(4186)(3)(40)=502320J##. This amount of heat will be absorbed initially by the ice to be brought to zero degrees and later to melt ##\frac{1}{3}## of the mass.

The heat absorbed by the ice for reach ##0## degrees can be calculated as: ##Q=cm(4 °C)=8360J##.

The heat absorbed by the ice in order to melt ##\frac{1}{3}## of the mass can be calculated with the ice latent heat of fusion as: ##\frac{3.33 x 10^5}{3}=111000J##
.
At this point, I can calculate the amount of total heat that the ice absorbs, which is the sum of ##8360J## and ##111000J## which is equal to ##119360J##, I can subtract it from the amount of total heat that the water gives up, and I get the amount of heat given up by the water that is not absorbed by the ice and must therefore be absorbed by the container. This quantity is ##502320J-119360J=382960J##.

I imagined that the initial temperature of the container was -4 degrees so this must absorb that amount of heat and have a heat capacity that allows it to raise its T by 4 degrees, absorbing that amount of heat.

From here I can calculate the final ##C=\frac{382960J}{4}=95740 J/K##

We are at the end of the problem, is this the right way of reasoning?

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• Salmone
Your reasoning looks good to me. Your calculations also look correct except for this subtraction:
Salmone said:
##502320J-119360J=383785J##.

• Salmone
TSny said:
Your reasoning looks good to me. Your calculations also look correct except for this subtraction:
Thank you, corrected.