A copper block with a mass of 300 grams is cooled to 77K 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 x 105 J/kg. When the system comes to equilibrium, the temperature is 0°C. (a) Calculate the maximum possible mass of water that could be in the cup. (g) (b) Calculate the minimum possible mass of water that could be in the cup. (g)

mass of copper block = 0.3 kg at 77 K or - 196 °C

Let mass of water taken be m (maximum), in this case we will have mixture of water and copper block at 0°C finally.

Heat gained by copper block

= msΔt

.3 x 196 x 385 J

heat lost by water to lower its temp fro 50°C to 0°C

= m x 4186 x 50

Heat lost = heat gained

m x 4186 x 50 =.3 x 196 x 385

m = 108.16 gm

Let mass of water taken be m (minimum), in this case we will have mixture of ice and copper block at 0°C finally.

heat lost by water to lower its temp fro 50°C to 0°C

= m x 4186 x 50

heat lost by water to lower its temp from 0°C to 0°C ice

= m x latent heat of fusion

m x 3.35 x 10⁵

Total heat lost

= m x 3.35 x 10⁵ + m x 4186 x 50

= 544300 m

Heat gained by copper block

= msΔt

.3 x 196 x 385 J

Heat lost = heat gained

544300 m =.3 x 196 x 385 J

m = 41.6 gm