Question

Water at ${80}^{\circ }C$is poured into a bucket containing $1.5kg$ of crushed ice at $0^{\circ }C$, such that all the ice melts and the final temperature recorded is $0^{\circ }C$. Calculate the amount of hot water added to the ice.[Take the specific heat capacity of water is $4200JK{g}^{-1}{K}^{-1}$ and the specific latent heat of ice $=336\times {10}^{3}jK{g}^{-1}$.]

Answer 1

Step 1: Given data

1. The quantity of crushed ice in the bucket is $1.5kg$
2. Take the specific heat capacity of water is $4200JK{g}^{-1}{K}^{-1}$ and the specific latent heat of ice $=336\times {10}^{3}JK{g}^{-1}$

Step 2: Calculating the heat given get by hot water

Let the amount of hot water need to melt all the ice is $xKg$.

Heat loss by hot water

$$\begin{gathered} =mc{\theta }_f \\ =x\times 4200\times (80-0) \\ =x\times 4200\times 80 \end{gathered}$$

Step 3: Calculating the heat absorbed by ice to form water at $0^{\circ }C$

Heat absorbed by ice for converting into the water at $0^{\circ }C$

$$\begin{gathered} =mL \\ =1.5\times 336\times {10}^3 \end{gathered}$$

Step 4: Calculating the amount of hot water added to the ice

By the principle of calorimetry,

Now heat absorbed by ice = heat produced by water

$$\begin{gathered} (x\times 4200\times 80)=1.5\times 336\times {10}^3 \\ \Rightarrow x=\frac{1.5\times 336\times {10}^3}{4200\times 80} \\ \Rightarrow x=1.5kg \end{gathered}$$

Hence, the amount of hot water requires $1.5kg$.