An open container holds ice of mass 0.595kg at a temperature of -13.5 celcius .

Question

An open container holds ice of mass 0.595kg at a temperature of -13.5 celcius . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 840 J/minute. The specific heat of ice to is 2100 J/kg and the heat of fusion for ice is 332 x 10^3 J/kg a) How much time passes before the ice starts to melt? (In minutes) b) From the time when the heating begins, how much time does it take before the temperature begins to rise above 0 degrees celcius? (in minutes) An open container holds ice of mass 0.595kg at a temperature of -13.5 celcius . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 840 J/minute. The specific heat of ice to is 2100 J/kg and the heat of fusion for ice is 332 x 10^3 J/kg a) How much time passes before the ice starts to melt? (In minutes) b) From the time when the heating begins, how much time does it take before the temperature begins to rise above 0 degrees celcius? (in minutes)

Explanation / Answer

(a) heat required so that ice starts melting = 0.595*2100*13.5=16868.25 J............................

time taken = 16868.25 / 840=20.081 min...........................................................................

(b)heat required to melt the ice = 0.595*332*10^3 = 197540 J................................................

time taken = 197540 / 840 = 235.166 min

hence total time from the beginning = 20.081 + 235.166 = 255.247 min

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