Question

15 g of water at 100°C is mixed with 15 g of ice at 0°C. What is the temperature of the mixture?

• A. 50° C
• B. 66.6°C
• C. 10°C
• D. 61.3°C

Answer 1

The correct option is C

10°C

Step 1:Given data

$15g$of water at ${100}^{0}C$

$15g$ of ice at $0^{0}C$

We know that the specific heat of water $c$ is $1cal{g}^{-1}$

and latent heat of fusion of ice $L$ is $80cal{g}^{-1}$

Step 2: Calculate heat released when the temperature of the water drops from ${100}^{0}C$ to $0^{0}C$

$Q_1=cm∆T$ where $m$ is the mass of water

$$\begin{gathered} Q_1=1\times 15\times (0-100) \\ {Q}_1=-1500cal \end{gathered}$$

Step 2: Heat required to melt ice at $0^{0}C$-

Energy absorbed by ice during the change in its physical state i.e. while melting without changing its temperature is called Latent heat.

since this energy is absorbed by the system. Hence it is positive. It is calculated as:

$Q_2=LM$ where $M$ is the mass of ice

$$\begin{gathered} Q_2=80\times 15 \\ {Q}_2=1200cal \end{gathered}$$

Step 3: Calculate total heat change-

$Q=Q_1+Q_2$

$=-1500+1200=300cal$

Step 4: Calculate the final temperature of the mixture-

Let the final temperature be $T_f$ The total mass of the system will be the sum of the mass of water and ice i.e. $M^t=30g$

Now $Q=c{M}^t(T_f-0)$

$$\begin{gathered} \Rightarrow 300=1\times 30T_f \\ \Rightarrow T_f={10}^{0}C \end{gathered}$$

Therefore option C is correct, the final temperature of the mixture is ${10}^{0}C$