How many calories of heat are required to raise the temperature of 4 kg of water
ID: 1465448 • Letter: H
Question
How many calories of heat are required to raise the temperature of 4 kg of water from 50 degrees F to the boiling point? A 5 gal container (approx. 20 kg) having a temperature of 212 degree F is added to a 50 gallon tub (approx. 200 kg) of water having a temperature of 50 degree F. What is the final equilibrium temperature (in degree C) of the mixture? A 5 kg piece of lead (specific heat 0.03 cal/g degree C) having a temperature of 80 C is added to 500 g of water having a temperature of 20 degree C What is the final equilibrium temperature (in degree C) of the system? A 300 g glass thermometer initially at 25 degree C is put into 200 cm^3 of hot water at 95 degree C. Find the final temperature (in degree C) of the thermometer, assuming no heat flows to the surroundings. (The specific heat of glass is 0.2 cal/g degree C). How much heat (in kcal) is needed to convert 1 kg of Ice into steam? How much heat (kcal) is needed to melt a 10 kg block of aluminum which has an Initial temperature of 30 degree C? (The specific heat, latent heat of fusion and melting point of aluminum are 0.215 cal/g degree C. 21.5 cal/g and 660 degree C.)Explanation / Answer
(1)
initial temperature = t1 = 10 oC
final temperature = t2 = 100 oC
Q = m*C*dT = 4*10^3*1*(100-10) = 3.6*10^5 cal
OPTION (b)
(2)
t1 = 212 = 100 oC
t2 = 50 oF = 10 oC
heat lost = heat gain
5*1*(100-t) = 50*1*(t-10)
t = 18 oC
(3)
t1 = 80 oC
t2 = 20 oC
heat lost by lead = heat gain by water
5*0.03*(80-t) = 0.5*1*(t-20)
t = 34 oC
OPTIOn d
(4)
heat lost by water = heat gained by glass
200*10^-6*1000*1*(95-t) = 0.3*0.2*(t-25)
t = 79
OPTION (d)
(5)
Q = m*(Lf + (Cw*100)+Lv)
Q = 1*(334*10^3 + (4190*100)+(2256*10^3)) = Q=3009000 J
Q = 720 kcal <<-answer
(6)
Q = m*(c*dT1 + L )
Q = 10*10^3*((0.215*(660-30)) + 21.5)
Q = 1570 kcal
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