Question

# Suppose you pour 0.250 kg of water at 20.0 ° C (about one cup) in a...

Suppose you pour 0.250 kg of water at 20.0 ° C (about one cup) in a 0.500 kg aluminum tray out of the oven at a temperature of 150 ° C. Suppose no transfer of heat occurs to anything else: the tray is placed on an insulated platform, and the transfer of heat to the air is neglected in the short time necessary to reach equilibrium. Also assume that a negligible amount of water boils. (a) What is the temperature when the water and the tray reach thermal equilibrium?

Mass of aluminium tray = ma = 0.5 kg

Mass of water = mw = 0.25 kg

Initial temperature of aluminium tray = T1 = 150 oC

Initial temperature of water = T2 = 20 oC

Temperature at thermal equilibrium = T

Specific heat of aluminium = Ca = 900 J/(kg.K)

Specific heat of water = Cw = 4186 J/(kg.K)

The heat lost by the aluminium tray is equal to the heat gained by the water.

maCa(T1 - T) = mwCw(T - T2)

(0.5)(900)(150 - T) = (0.25)(4186)(T - 20)

67500 - 450T = 1046.5T - 20930

1496.5T = 88430

T = 59.09 oC

The temperature when the water and the tray reach equilibrium = 59.09 oC