Question #1635

A 500g lead mass is heated to 150 C and placed on a block of ice at -10 C. How much ice, if any will melt?

Expert's answer

Specific heat of the lead C_{l}= 130 J/kg*K

The heat equation is

C_{l}(T_{2}-T_{1})m_{l} = C_{ice}(T_{3}-T_{1})m_{ice} + C_{fice}m_{ice}

T1=0C, T2 = 150C, T3= -10C, ml = 0.5 kg:

130 [J/kg K]*150K*0.5 kg = 2.11x10^{3}[J/kgK]*10[K] m+ 333.55x10^{3}[J/kg] m

9750 = 21100m+333550m

m = 0.1784 kg = 178.4 g of ice.

The heat equation is

C

T1=0C, T2 = 150C, T3= -10C, ml = 0.5 kg:

130 [J/kg K]*150K*0.5 kg = 2.11x10

9750 = 21100m+333550m

m = 0.1784 kg = 178.4 g of ice.

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