A piece of copper of mass 20g at a temperature of 110°C was dropped into a mixture of ice and water at 0°C. If the final steady temperature of the mixture is 0°C . Calculate the amount of ice that melted [Specific heat capacity of copper = 0.4 Jg\(^{-1}\)K\(^{-1}\), specific latent heat of fusion of ice = 330Jg\(^{-1}\)]
- 0.37g
- 0.60g
- 2.40g
-
2.70g
The correct answer is: D
Explanation
Mice = \(\frac{MC}{\theta}\)
= \(\frac{20 \times 0.4 \times 110}{330}
Heat lost by copper = Heat gained by ice
m\(_{copper}\) c \(\theta\) = m\(_{ice}\) L
20 x 0.4 x (110 - 0) = m\(_{ice}\) x 330
m\(_{ice}\) = \(\frac{20 \times 0.4 \times 110}{330}\)
= \(\frac{8}{3}\)
= 2.67 g
= 2.70g (to 1 d.p)