Question

How many gram of ice at $-{10}^{°}\mathrm{c}$ is needed to cool $100\mathrm{g}$ of water from ${25}^{°}\mathrm{c}$ to ${10}^{°}\mathrm{c}$?

Answer 1

Step 1: Given data

• Mass of water(m)=$100\mathrm{g}$
• Latent heat of fusion of ice($∆\mathrm{H}$)= $80$
• Specific heat of ice($\mathrm{s}$)= $0.5$
• Temperature ($∆\mathrm{T}$)=${25}^{°}\mathrm{c}-{10}^{°}\mathrm{c}$$={15}^{°}\mathrm{c}$

Step 2: Calculation of Heat change for hot body and cold body

Formula for calculation of heat change(Q)=$\mathrm{m}\times \mathrm{s}\times ∆\mathrm{t}$

Putting the above values Q(Hot Body)=$100\times 1\times 15=1500$

Q(Cold body)=

$$\begin{gathered} \mathrm{m}\left(\mathrm{ice}\right)\times 0.5\times 10+\mathrm{m}\left(\mathrm{ice}\right)\times 80+\mathrm{m}\left(\mathrm{ice}\right)\times 1\times 10 \\ =95\mathrm{m}\left(\mathrm{ice}\right) \end{gathered}$$

Step 3: Calculation of mass of Ice

As in Thermodynamics, Q(Hot body)=Q(Cold body)

Putting the both values of Q(Hot body) and Q(Cold body) in the above equation:

$$\begin{gathered} 95\mathrm{m}\left(\mathrm{ice}\right)=1500 \\ \mathrm{m}\left(\mathrm{ice}\right)=\frac{1500}{95} \\ \mathrm{m}\left(\mathrm{ice}\right)=15.78\mathrm{g} \end{gathered}$$

so, mass of ice= $15.78\mathrm{g}$

Therefore, mass of ice that is needed to cool is $15.78\mathrm{g}$.