Answer to Question #1635 in Molecular Physics | Thermodynamics for Bila Hussa
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?
Specific heat of the lead Cl= 130 J/kg*K
The heat equation is
Cl(T2-T1)ml = Cice(T3-T1)mice + Cficemice
T1=0C, T2 = 150C, T3= -10C, ml = 0.5 kg:
130 [J/kg K]*150K*0.5 kg = 2.11x103[J/kgK]*10[K] m+ 333.55x103[J/kg] m
9750 = 21100m+333550m
m = 0.1784 kg = 178.4 g of ice.