An Example on Heat Lost/Gained in a System


Problem:

There is 8000 g of water in a 1500 g bucket. The initial temperatures of the water and bucket are 75 ° C and 20 ° C respectively. The heat capacity of water is 4186 J/kg K and the heat capacity of the bucket is 1000 J/kg K. Determine the final temperature of the water/bucket system.

Solution:

First we must determine what is going on in this problem. There is heat that is being transferred. The heat moves from the water to the bucket since the water is at a higher temperature and heat flows from hot to cold. The temperature of the water will decrease as more heat is transferred while the temperature of the bucket will increase as more heat is transferred. Eventually, the water and the bucket will reach the same final temperature. Even though heat is still being transferred, it is transferred equally now, from the water to the bucket and vice versa. In order to determine this final temperature, the following equation must be used:

Q = m Cp D T

Recall that Q is the amount of heat lost or gained. We want to know the final temperature of the system, and we have determined that occurs when the amount of heat transferred from the water to the bucket equals the amount of heat that is transferred from the bucket to the water. We will set Q for the water to the bucket equal to Q for the bucket to the water.

-Qw = Qb

where Qw = heat transferred from the water to the bucket

          Qb = heat transferred from the bucket to the water

Since heat is lost from the water and gained by the bucket, the sign for Qw is negative and the sign for Qb is positive.

Plugging in the definition of Q gives the following equation

-mwCpw(TF-T1,w) = mbCpb (TF-T1,b)

where mw = mass of the water

          Cpw = heat capacity of the water

          mb = mass of the bucket

          Cpb = heat capacity of the bucket

          T1,w = initial temperature of the water

          T1,b = initial temperature of the bucket

          TF = the final temperature of the system

All of the variables in this equation are known from the problem statement, except the final temperature (TF). This is the variable that we want to solve for. Solving the above equation for TF gives

TF = (-mwCpwT1,w mbCpbT 1,b)

----------------------------------

(-mwCpw mbCpb)

From the problem statement

          mw = 8000 g

          mb = 1500 g

          Cpw = 4186J/kg K

          Cpb = 1000J/kg K

          T1,w = 75 ° C

          T1,b = 20 ° C

Some of these numbers need to be converted into different units. The temperatures given need to be in Kelvin, not ° C. The grams must also be converted into kilograms. You can go to the Automatic Unit Converter to convert to the correct units. Using the converter, the new temperatures and masses are

          T1,w = 348 K

          T1,b = 293 K

          mw = 0.800 kg

          mb = 0.150 kg

Plugging in all these numbers yields:

TF = [-(0.800 kg)*(4186J/kg K)*(348 K) (0.150 kg)*(1000J/kg K)*(293 K)]

      ------------------------------------------------------------------------------------

[-(0.800 kg)*(4186J/kg K) (0.150 kg)*(1000J/kg K)]

TF = 345 K

We have just shown how to calculate the final temperature of the water/bucket system!


Proceed to Cooking a Potato
Proceed to Heat Transfer Between a Plate and Your Food
Return to Heat Lost/Gained by a System
Return to Heat Transfer
[Introduction | Kinetics | Heat Transfer | Mass Transfer | Bibliography]

This project was funded in part by the National Science Foundation and is advised by Dr. Masel and Dr. Blowers at the University of Illinois.

2007 Arizona Board of Regents for The University of Arizona