Question

The water of mass 75g at 100°C is added to ice of mass 20g at -15°C.What is the resulting temperature.(Latent heat of ice=80cal/g and specific heat of ice=0.5)approx?

Answer 1

Find the total heat used by ice in changing state from solid to liquid(water)
Q1 = 20g*80cal/g = 1600 cal

1600 cal = 6694.4 J = 6.69 kJ
M'C(T'-x) = MC(x-T) + 1600
where,
M'=Mass of water = 75 g
M=mass of ice= 20g
C= specific heat of water = 4.18 J/g·°C = 0.00418 kJ/g·°C
T' = Temp of water = 100 °C
T= Temp of ice = -15 °C
x= final temp
75 * 0.00418 * (100- x) = 20 * 0.00418 * (x+15) + 6.69
31.4 – 0.314x = 0.084x + 1.25 + 6.69

-0.398x = -23.46

x = 60 °C

The resulting temperature = 60 °C