An Example: Heat Transfer of the Steak/Plate System

Now that we have described how to determine the heat lost or gained by a system, and have done a general example on how to use this equation, it is now time to do a specific example. This example will use the equation discussed on the previous page to determine the final temperature of the steak/plate system.

A 260 g steak is resting on a 200 g glass plate. The initial temperature of the steak is 85 ° C and the initial temperature of the plate is 30 ° C. The heat capacity of the steak is approximately 0.5 J/kg K and the heat capacity of the glass plate is 1.2 J/kg K. Determine the final temperature of the steak/plate system.

First we must determine what is going on in this problem. There is heat that is being transferred. The heat moves from the steak to the plate since the steak is at a higher temperature and heat flows from hot to cold. The temperature of the steak will decrease as more heat is transferred while the temperature of the plate will increase as more heat is transferred. Eventually, the steak and the plate will reach the same final temperature. Even though heat is still being transferred, it is transferred equally now, from the steak to the plate 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 steak to the plate equals the amount of heat that is transferred from the plate to the steak. We will set Q for the steak equal to Q for the plate.

-Q_s = Q_p

where Q_s = heat transferred from the steak to the plate

          Q_p = heat transferred from the plate to the steak

Since heat is lost from the steak and gained by the plate, the sign for Q_s is negative and the sign for Q_p is positive.

Plugging in the definition of Q gives the following equation

-m_sCp_s(T_F-T_1,s) = m_pCp_p (T_F-T_1,p)

where m_s = mass of the steak

         Cp_s = heat capacity of the steak

         m_p = mass of the plate

         Cp_p = heat capacity of the plate

         T_1,s = initial temperature of the steak

         T_1,p = initial temperature of the plate

         T_F = the final temperature of the system

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

T_F = (-m_sCp_sT_1,s - m_pCp_p T_1,p)

           (-m_sCp_s – m_pCp_p)

From the problem statement

         m_s = 260 g

         m_p = 200 g

         Cp_s = 0.5 J/kg K

         Cp_p = 1.2 J/kg K

         T_1,s = 85 ° C

         T_1,p = 30 ° 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

         T_1,s = 358 K

         T_1,p = 303 K

         m_s = 0.260 kg

         m_p = 0.200 kg

Plugging in all these numbers yields:

T_F = [-(0.260 kg)*(0.5J/kg K)*(358 K) – (0.200 kg)*(1.2J/kg K)*(303 K)]

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

[-(0.260 kg)*(0.5J/kg K) – (0.200 kg)*(1.2J/kg K)]

We have just shown how to calculate the final temperature of a system, let’s think about what else we can do with the steak/plate system. Let’s determine the heat that was transferred from the steak into the plate. We can do this now that we know the final temperature of the system.

Problem:

Determine the heat transferred between a 260 g steak resting on a 200 g plate. The thermal conductivity of the steak is 0.49 W/m K and the area the steak covers on the plate is 10 cm². The thickness of the steak is 3.5 cm. Again the initial temperature of the steak and plate are 85 ° C and 30 ° C respectively.

From the calculation we did above, we know the final temperature of the system. We can also use the equations that describe conductive heat transfer to determine how much heat was transferred to the plate from the steak. Recall that the equation describing conduction is

Q = -kA(D T)/( D x)

where Q = heat conducted

          k = thermal conductivity

          A = surface area for conduction

          D T = T_F – T_i

          D x = thickness

All of the variables in the above equation are known except Q which is what we want to solve for. The numbers given in the problem statement are:

          k = 0.49 W/m K

          A = 10 cm²

          D T = T_F – T_i = 358 K – 322 K = 36 K

          D x = 3.5 cm

Some of these numbers again have to be converted. Using the Automatic Unit Converter to change the area from cm² to m² and the thickness from cm to m gives the following numbers:

          A = 0.001 m²

          D x = 0.035 m

Plugging in all these numbers into the heat conduction equation gives the following:

Q = -[(0.49W/m K)*(0.001 m²)*(36 K)]/(0.035 m)

This is how much heat the steak lost to the plate. Now you know how to use the equations for conduction and, in general, heat transfer!