An open container holds ice of mass 0.540\ m{kg} at a temperature of -11.7\ m{^{

Question

An open container holds ice of mass 0.540 m{kg} at a temperature of -11.7 m{^{circ} C} . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 790 m{J/minute} . The specific heat of ice to is 2100 { m J/kg cdot K} and the heat of fusion for ice is 334 imes 10^3;{ m J/kg}

Explanation / Answer

1)Heat energy required to bring the temperature of ice to 0*C, By Q = m x c x delta t =>Q = 0.515 x 2100 x 17.8 = 19250.7 J Time(t1) = 19250.7/900 = 21.39 min 2) Let the heat energy required to melt the ice is Q, =>Q = m x L = 0.515 x 334000 = 172010 J Time(t2) = 172010/900 = 191.12 min Thus the time from when the heating begins, it take before the temperature begins to rise above 0 degrees C = t1+t2 = 21.39+191.12 = 212.51 min

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