How much heat (in kcal) needs to be extracted from 2kg of water at 20 degree C t
ID: 1466022 • Letter: H
Question
How much heat (in kcal) needs to be extracted from 2kg of water at 20 degree C to make ice at -10 degrees C? ( the specific heat of ice is 0.5 cal/g degree C)?
A home is built on a concrete slab 20 cm thick. what is the approximate rate of heat transfer per area in (cal/s m^2) if the temperature in the home is 68 degree F and the ground temperature is 50 degree F ( the thermal conductivity of concrete is 2x10^-3 cal/s cm degree C)?
a 100 kg student eats a 200 calorie doughnut. To "burn it off", he decides to climb the steps of atall building. How high (in m) would he have to climb to expend an equilavent amount of work? (1 food calorie=10^3 calories)?
Explanation / Answer
heat extracted = Q = m*[ Cice*dT1 + Lf + Cw*dT2 ]
Q = 2*( (2100*10) +(334*10^3)+(4190*20) )
Q = 877600 J
Q = 877600*(1/4.2) = 208 kcal
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heat transfer rate H = k*A*dT/t
heat transfer per area = H/A = K*dT/t
dT = temperature difference = 68-50 = 18
t = thickness = 0.2 m
heat transfer per area = (2*10^-3*18)/20 = 0.0018 cal /cm^2 = 0.0018*10^4 cal /m^2 = 18 cal/m^2
++++++
work = m*g*h
200*4.2 = 100*9.8*h
h = 0.857 m
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