Question

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?

Answer 1

Mass of aluminium tray = m_a = 0.5 kg

Mass of water = m_w = 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 = C_a = 900 J/(kg.K)

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

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

m_aC_a(T₁ - T) = m_wC_w(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