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 ^{o}C

T_{i} = initial temperature ice poured = - 6
^{o}C

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

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