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

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

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.

_{l}= 130 J/kg*KThe 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] m9750 = 21100m+333550m

m = 0.1784 kg = 178.4 g of ice.

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment