Question

Some small aluminium rivets of total mass 170g and at 100°C are empitied into a hole a largert block of ice at 0°C. How much ice will melt?

Answer 1

The aluminium rivets will cooled while the ice will melted. Therefore, we can write the heat balance equation:

$$m_{al}c_{al}\Delta T=m_{ice}L_f,$$ $$m_{ice}=\dfrac{m_{al}c_{al}\Delta T}{L_f},$$ m_{ice}=\dfrac{0.17\ kg\cdot930\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot100\ ^{\circ}C}{3.33\cdot10^5\ J}=0.047\ kg.

Answer:

$m_{ice}=0.047\ kg.$