Question

An open container holds ice of mass 0.565kg at a temperature of -13.5?C . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 740J/minute .

The specific heat of ice to is 2100 J/kg?K and the heat of fusion for ice is 334

Answer 1

  

\(Q=mc\Delta T\) , where Q is the heat energy, m is the mass, c is the specific heat capacity and    is the change in temperature.

A) We know that ice melts (or water freezes) at 0C, so we first need to calculate how much energy it takes for the ice to reach this temperature. Using the above equation, with a temperature change of 13.5C, we find a value for Q of 16017 J.

Knowing that we're supplying the heat at a rate of 740 J/min, we divide the total heat by this to find a time of 21.6 minutes, or 21 min 39 seconds, if you prefer.

B) The temperature will not rise above 0C until all the ice has melted, as energy is now going into melting the ice. The energy to melt all the ice is simply the heat of fusion (or melting) per kilogram, multiplied by the number of kilograms we have; 0.565 x 334 x10^3 = 188710 J.

Again dividing by 740 we find a time of 255.01 min.

Since it's the time from when we started heating the question wants, we add the two times to give 276.6 min, or 4 hours 37 minutes.