Question

4. An iron container has a mass of 200 g and contains 50 g of water 40 C. 50 g of ice are poured at -6° C. Calculate the equilibrium temperature and describe the final composition.

5. We have a puddle of hot water @ 60 ° C exposed to air @ 28 ° C. Assume that the cooling constant of the water is 0.05/minutes. Calculate how long it will take the temperature on the surface of the water to go down to 42° C.

Answer 1

4.

m_i = mass of iron container = 200 g = 0.2 kg

m_w = mass of water inside the container = 50 g = 0.050 kg

m = mass of ice poured = 50 g = 0.050 kg

T_o = initial temperature of iron container and water = 40 ^oC

T_i = initial temperature ice poured = - 6 ^oC

T_f = final equilibrium temperature = ?

c_i = specific heat of iron = 0.450 J/gC

c_w = specific heat of water = 4.2 J/gC

c = specific heat of ice = 2.11 J/gC

m' = mass of ice melted

using conservation of heat

heat lost by iron container + heat lost by water = heat gained by ice

m_i c_i (T_o - T_f ) + m_w c_w (T_o - T_f ) = m c (T_f - T_i) + m L

(0.2) (0.450) (40 - 0) + (0.050) (4.2) (40 - 0) = (0.050) (2.11) (0- (- 6)) + m' (334)

m' = 0.034 kg = 34 g

hence water = 34 + 50 = 84 g

ice = 50 g - 34 g = 16 g

final equilibrium temperature comes out to be 0 C since the heat of melting of the ice is greater than the heat lost by the iron container and water