Question

a) An ice cube initially at zero degrees C has mass 37g. It is used to cool a hot cup of water at 184 which has mass 200g. What is the final temperature in degrees Celcius? Use as specific heat of water 4.18 kJ/(kg K) and the heat of fusion 334 kJ/kg.

b) A mass of 1kg is attached to a spring with spring constant 10N/m. How many meters down does its equilibrium position fall when it is used to hang the mass vertically?

Answer 1

(a) Given the mass of ice mi = 37g, the initial temperature of ice is Ti = 0C, mass of water mw = 200g and the initial temperature of water Tw = 184C. Also the specific heat of water cw = 4.18kJ/(kg K) and the heat of fudsion of ice Lf = 334 kJ/kg.

Let the final temperature of the mixture be TC. To get to the temperature of TC, 200g of water at 184C will release heat. The amount of heat released by 200g of water is given by

The 37g ice at 0C will absorb this heat and first melt to 37g water at 0C and then to 37g water at TC.

The amount of heat absorbed by 37g of ice at 0C to melt into 37g of water at 0C is given by

The amount of heat absorbed by the 37g of water at 0C to become 37g of water at TC is given by

So the total heat absorbed by the ice is

By principle of method of mixtures, the heat lost by the hot body is equal to the heat gained by the cold body. Therefore

The final temperature of the mixture is 142.8C.

(b) Here the mass attached to the spring is 1kg and the spring constant is k = 10N/m. Let x be the extension of the spring from the equilibrium position when the mass is released. Now the forces are equal and opposite. Therefore

Spring force = Weight of mass

So the equilibrium position will fall 0.196m downwards.