Two thermal reservoirs (\"hot\", TH = 134°C and \"cold\", TC = 9.0°C) are put in
ID: 1785500 • Letter: T
Question
Two thermal reservoirs ("hot", TH = 134°C and "cold", TC = 9.0°C) are put into thermal contact by a lead slab. The slab's length is L = 23.0 cm and its cross-sectional area A = 94.0 cm2. a.) Find the initial rate of heat conduction for each reservoir when they are connected. PH = J/s PC = J/s slab prob 1 b.) If a given reservoir does not appreciably change temperature with heat flow, it could be because (select all that apply): it is undergoing phase change it has a very large mass it has a very large specific heat capacity c.) Assuming one/both reservoirs does change temperature with heat flow, other approximations that will improve the accuracy of a subsequent calculation are: the specific heat capacity of the conducting medium is that of the reservoirs the thermal conductivity of the conducting medium is that of the reservoir
Explanation / Answer
a) length of lead block ,l = 23 X 10^(-2) mrs ; area of cross-section = 94 X 10^(-4) m^2
the thermal conductivity constant of lead = 35 watts/s /mr K
Rate of thermal conduction = 35 J/s /mr K X 94 X 10^(-4) m^2 X (134-9)/ [23 X 10 ^(-2)m]
= 178.8 J/s
b) The heat transfer adds to heat energy to the resorvior at lower temperature. If inspite of addition of heat energy ,the temperature of the colder resorvoir does not change , it could be that the energy is absorbed for changing the state of resorvoir material. For example if colder resorvoir has ice, the heat tranfered is absorbed in melting process. Similarly if the mass of resorvoir is very large , the heat added may not raise the temperature considerably. If the specific heat of colder material is very high amount of heat required to raise temperature of 1 Kg of substance by 1 degree will be very high..
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