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which equation do i use?
An 800g rubber hot water bottle at 15 degrees is filled with 1.50kg of water at a temperature of 80 degrees. Before being placed in a cold bed, thermal equilibrium is reached between the bottle and the water.
A - What is the common temperature of the rubber and water at this time? (Note: the specific heat capacity of rubber is 1700Jkg-1K-1)
B - the hot water bottle is placed in the cold bed when the temperature of the bottle is 65 degrees. Eight hours later its temperature is measured to be 15 degrees. At what average rate has it lost its internal energy?
This is just conservation of energy. You have to use the facts that (1) the water bottle and water end up at the same temperature (let's call it T1), and (2) all the heat going into the bottle comes from the hot water. Then you can use Q = m * c * θ to work out the amount of heat in each case.
A: Heat gained by hot water bottle = 0.8 * 1700 * (T1 - 15) = 1360 * (T1 - 15).
Heat lost by water = 1.5 * 4200 * (80 - T1) = 6300 * (80 - T1).
1360 * (T1 - 15) = 6300 * (80 - T1).
1360 * T1 - 20400 = 504000 - 6300 * T1.
7660 * T1 = 524400
T1 = 68.5°C.
B: The full hot water bottle has a combined heat capacity of 0.8 * 1700 + 1.5 * 4200 = 1360 + 6300 = 7660 J.K-1.
Over 8 * 3600 seconds it becomes 50°C cooler.
So total amount of heat lost = 7660 * 50 = 383000J.
Rate of heat loss = 383000 / (8 * 3600) = 383000 / 28800 = 13.3 Watts.