Question

$200\text{g}$ of ice at $-{20}^{\circ }\text{C}$ is mixed with $500\text{g}$ of water at ${20}^{\circ }\text{C}$ in an insulating vessel. Final mass of the water in the vessel is -
(specific heat of ice $=0.5{\text{cal g}}^{-1}{{}^{\circ }\text{C}}^{-1}$ & specific heat of water $=1{\text{cal g}}^{-1}{{}^{\circ }\text{C}}^{-1}$)

• A. $700\text{g}$
• B. $600\text{g}$
• C. $500\text{g}$
• D. $200\text{g}$

Answer 1

The correct option is B $600\text{g}$

Given
Mass of water $\left(m_1\right)=500\text{g}$
Mass of Ice $\left(m_2\right)=200\text{g}$
Initial temperature of water $\left(T_1\right)={20}^{\circ }\text{C}$
Initial temperature of ice $\left(T_2\right)=-{20}^{\circ }\text{C}$
Maximum heat lost by water to reach the temperature of $0^{\circ }\text{C}$
$Q_1=500\times 1\times (20-0)$
$=10,000\text{cal}$
Heat required to raise the temperature of ice upto $0^{\circ }\text{C}$
$Q_2=200\times 0.5\times 20$
$=2000\text{cal}$
$\Rightarrow Q_1-Q_2=8000\text{cal}$
The remaining heat will be used to just melt the ice into water at $0^{\circ }\text{C}$.
But we know that the energy required to just melt $m\text{grams}$ of ice is given by $Q=m\times L$
Thus, we can say from the data obtained that $8000=m\times 80$
$\Rightarrow$ Mass of ice melted into water $\left(m\right)=100\text{g}$
Hence, total mass of water at $0^{\circ }\text{C}$ in the vessel will be
$M=m_1+m=500+100=600\text{g}$
Thus, option (b) is the correct answer.