JAMB - Physics (1999 - No. 15)
Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss.
Specific heat capacity of copper = 400JKg-1K-1
Latent heat of fusion of ice = 3.3 x 10\(^5\)JKg-1
Specific heat capacity of copper = 400JKg-1K-1
Latent heat of fusion of ice = 3.3 x 10\(^5\)JKg-1
8/33kg
33/80kg
80/33kg
33/8kg
Εξήγηση
Heat given off by the copper = MCt
= 2 x 400 x (100 - 0)
Heat absorbed by the ice = ML
= M x 3.3 x 10\(^5\)
Heat lost = Heat gained
∴ M x 3.3 x 10\(^5\) = 2 x 400 x 100
∴ M =\(\frac{(2 \times 400 \times 100)}{3.3 \times 10^5}\)
= \(\frac{8}{33}\)Kg
= 2 x 400 x (100 - 0)
Heat absorbed by the ice = ML
= M x 3.3 x 10\(^5\)
Heat lost = Heat gained
∴ M x 3.3 x 10\(^5\) = 2 x 400 x 100
∴ M =\(\frac{(2 \times 400 \times 100)}{3.3 \times 10^5}\)
= \(\frac{8}{33}\)Kg