Question

An insulated thermos contains 124.0 cm³ of hot coffee at a temperature of 80.0 °C. You put in 12.0 g of ice cube at its melting point to cool the coffee. What is the temperature of the coffee once the ice has melted and the system is in thermal equilibrium? Treat the coffee as though it were pure water. (Answer in °C)

Answer 1

Here we have -

124 cm³ = 0.124 kg
12g = 0.012 kg
specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
specific heat of steam is 2.1 kJ/kgK
heat of fusion of ice is 334 kJ/kg

Now, energy required to melt the ice and warm it to temperature T
E1 = [ 334 kJ/kg x 0.012 kg ] + [ 4.186 kJ/kgC x 0.012 kg x (T–0)K ]
E2 = 4.186 kJ/kgC x 0.124 kg x (80–T)K

Set them equal and solve for T -

4.008 + 0.050 = 0.519*(80 - T)

=> 80 - T = 7.82

=> T = 80 - 7.82 = 72.18 deg C.

So, the temperature of the coffee after thermal equilibtium = 72.18 deg C.