Question

$10\text{g}$ of ice at $-{20}^{\circ }\text{C}$ is added to $20\text{g}$ of water at ${20}^{\circ }\text{C}$. Assuming that no heat is lost to the surrounding, find the final temperature of the mixture.
[Take, specific heat capacity of ice $C_i=2100{\text{J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}$, specific heat capacity of water $C_w=4200{\text{J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}$ & Latent heat of fusion of ice $L=3.36\times {10}^5{\text{J kg}}^{-1}$]

• A. $0^{\circ }\text{C}$
• B. $1^{\circ }\text{C}$
• C. $2^{\circ }\text{C}$
• D. $3^{\circ }\text{C}$

Answer 1

The correct option is A $0^{\circ }\text{C}$

The heat released by water when it cools down from ${20}^{\circ }\text{C}$ to $0^{\circ }\text{C}$ is given by :
$Q_1=m{C}_w\Delta T$
$Q_1=0.02\text{kg}\times 4200{\text{J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}\times {20}^{\circ }\text{C}=1680\text{J}....\left(1\right)$
The heat absorbed by ice to reach $0^{\circ }\text{C}$ from $-{20}^{\circ }\text{C}$ is given by:
$Q_2=m^{\prime }{C}_i\Delta T$
$Q_2=0.01\text{kg}\times 2100{\text{J kg}}^{-1}{}^{\circ }{\text{C}}^{-1}\times {20}^{\circ }\text{C}=420\text{J}.....\left(2\right)$
Also:
Heat required to just melt $10\text{g}$ of ice is given by :
$Q_3=m^{\prime }L=0.01\text{kg}\times 3.36\times {10}^5{\text{J kg}}^{-1}=3360\text{J}.....\left(3\right)$

From the principle of calorimetry,
Heat lost by water = Heat gained by ice.
$Q_1=Q_2+Q_3$
But from $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$, we can say that,
$Q_1<Q_2+Q_3$
Also, $Q_1<Q_3$ and $Q_1>Q_2$
Thus, we can conclude that all the ice will not be melt and temperature becomes $0^{\circ }\text{C}$.
Hence, option (a) is the correct answer.