At time t = 0, a vessel contains a mixture of 10 kg of water and an unknown mass

Question

At time t = 0, a vessel contains a mixture of 10 kg of water and an unknown mass of ice in equilibrium at 0 degrees C. The temperature of the mixture is measured over a period of an hour, with the following results: During the first 50 min, the mixture remains at 0 degrees C; from 50 min to 60 min, the temperature increases steadily from 0 degrees C to 2 degrees C. Neglecting the heat capacity of the vessel, determine the mass of ice that was initially placed in it. Assume constant power input to the container.

Explanation / Answer

Let x be the constant power factor per minute of time. And let m be the mass in Kg of ice. Then during the 50 minutes the constant power does the work to convert the zero degree ice to 0 degree water. The work done is converted into the heat energy mL, where m is the mass of ice and L is the latent heat of fusion of ice to water per Kg of ice. The work for 50 minutes by the power is done is 50x. So mL = 50x, where x could be treated as constant of proportion. From 50 to 60 degree, the constant power of x per minute does a work of 10x which raises the 10+m kg of water by 2 degrees = (10+m)*2. Using C= 1cal/kg/deg c Therefore mL/50 =x = (10+m)*2/10 mL=(10+m)10. =>mL=100+10m =>mL-10m = 100. =>(L-10)m = 100. =>m = 140/(L-10) = 100/(80-10) = 100/70 = 1.43kg. Here we used fact of latent heat of fusion of ice from ice to water = L =80 K cals So the mass of the ice in the experiment = 1.43kg

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