# 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
Thread moved from the technical forums to the schoolwork forums
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?

Last edited:

Last edited:
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.

## What is the first law of thermodynamics and how does it apply to heat transfer problems?

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transferred or converted from one form to another. In heat transfer problems, this means that the amount of heat lost by one system must equal the amount of heat gained by another system, assuming no energy is lost to the surroundings.

## What are the different modes of heat transfer?

There are three primary modes of heat transfer: conduction, convection, and radiation. Conduction is the transfer of heat through a solid material, convection is the transfer of heat through a fluid (liquid or gas) due to the fluid's movement, and radiation is the transfer of heat through electromagnetic waves without the need for a medium.

## How do you calculate the heat transfer through conduction?

The heat transfer through conduction can be calculated using Fourier's Law, which is given by the equation: $$Q = -kA \frac{dT}{dx}$$, where $$Q$$ is the heat transfer rate, $$k$$ is the thermal conductivity of the material, $$A$$ is the cross-sectional area through which heat is transferred, and $$\frac{dT}{dx}$$ is the temperature gradient.

## What is the difference between steady-state and transient heat transfer?

Steady-state heat transfer occurs when the temperature distribution in a system does not change with time, meaning the heat entering and leaving the system is constant. Transient heat transfer, on the other hand, occurs when the temperature distribution within the system changes with time, meaning the heat entering and leaving the system varies over time.

## How do you approach solving complex heat transfer problems?

To solve complex heat transfer problems, it is essential to follow a systematic approach: (1) Define the problem and identify the known and unknown variables. (2) Determine the mode(s) of heat transfer involved. (3) Apply the relevant equations and boundary conditions. (4) Simplify the equations if possible and solve for the unknowns. (5) Validate the results by checking for consistency and accuracy with physical principles and known data.

• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
950
• Introductory Physics Homework Help
Replies
3
Views
364
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
12
Views
1K