Question

A copper block with a mass of 500 grams is cooled to 77 K by being immersed in liquid nitrogen. The block is then placed in a Styrofoam cup containing some water that is initially at +50.0°C. Assume no heat is transferred to the cup or the surroundings. The specific heat of liquid water is 4186 J/(kg °C), of solid water is 2060 J/(kg °C), and of copper is 385 J/(kg °C). The latent heat of fusion of water is 3.35 * 10^5

J/kg. What is the mass of water in the cup, if the final temperature is −10.0°C?

g

Answer 1

heat released by water to get at -10 degree C

= m x 4186 x 50 + m x 3.35 x 10^5 + m x 2060 x 10

heat gained by copper block= 0.500 x 385 x (263 - 77) = 35805 J

heat released = heat gained ( from energy conservation)

m x 4186 x 50 + m x 3.35 x 10^5 + m x 2060 x 10 = 35805

m = 0.06338 kg = 63.4 gm

Hence,

The mass of water in the cup is 63.4 grams