

An Example on Heat Lost/Gained in a SystemProblem: 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. Q_{w} = Q_{b} where Q_{w} = heat transferred from the water to the bucket Q_{b} = heat transferred from the bucket to the waterSince 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_{w}Cp_{w}(T_{F}T_{1,w}) = m_{b}Cp_{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 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_{w}Cp_{w}T_{1,w} – m_{b}Cp_{b}T_{ 1,b})  (m_{w}Cp_{w} m_{b}Cp_{b}) 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 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)]  [(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. 
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