Answer to Question #1635 in Molecular Physics | Thermodynamics for Bila Hussa

Question

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?

Answer 1

Specific heat of the lead C_l= 130 J/kg*K
The heat equation is
C_l(T₂-T₁)m_l = C_ice(T₃-T₁)m_ice + C_ficem_ice

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.