Q Lost Q Gained

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.

I have been on keto for almost 4 weeks. I initially lost 4 lbs then gained 1 back. Now i lose 1 one day and gain it back the next. This keeps happening and has for almost a week. 6.41 Plan: Heat gained by water plus heat lost by copper tubing must equal zero, so q water = - q copper. However, some heat will be lost to the insulated container.

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.

-Q_w = Q_b

where Q_w = heat transferred from the water to the bucket

Q_b = heat transferred from the bucket to the water

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

Plugging in the definition of Q gives the following equation

-m_wCp_w(T_F-T_1,w) = m_bCp_b (T_F-T_1,b)

where m_w = mass of the water

Cp_w = heat capacity of the water

m_b = mass of the bucket

Qlost Qgained

Cp_b = heat capacity of the bucket

T_1,w = initial temperature of the water

T_1,b = initial temperature of the bucket

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_wCp_wT_1,w – m_bCp_bT _1,b)

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(-m_wCp_w m_bCp_b)

Measurements Of Heat

From the problem statement

m_w = 8000 g

m_b = 1500 g

Cp_w = 4186J/kg K

Cp_b = 1000J/kg K

T_1,w = 75 ° C

T_1,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

T_1,w = 348 K

Measuring Specific Heat

T_1,b = 293 K

m_w = 0.800 kg

m_b = 0.150 kg

Plugging in all these numbers yields:

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

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

T_F = 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.