Question

Answer

Q(c). The temperature of 170g of water at 50°C is lowered to 5°C by adding certain amount of ice to it. Find the mass of ice added.
Given: Specific heat capacity of water = 4200 J kg^–1°C^–1 and specific latent heat of ice =336000 J kg^–1

Answer 1

Dear Student ,
Let the mass of the ice be m kg .

$$\begin{gathered} \left(m\times 336000\right)+\left[0.170\times 4200\times \left(5-0\right)\right]=0.170\times 4200\times \left(50-5\right) \\ \Rightarrow m=\frac{0.170\times 4200\times 40}{336000}=0.085Kg=85gm \end{gathered}$$
Regards