Answer: Option A
:
A
The heat, which $0.15kg$ of water can release when its temperature is changed from ${20}^{\circ }C$ to $0^{\circ }C$.
$Q_1=m_W{s}_W\Delta T_W$
where, $m_W$ is the mass of the water, $s_W$ is the specific heat of water and $\Delta T_W$ is the change in temperature of water.
Given, $m_W=0.15kg$, $s_W=1kcalk{g}^{-1}$.
$Q_1=0.15\times (1\times {10}^3)\times (20-0)$
$=3000cal$ ...$\left(1\right)$
Now, heat absorbed by $0.10kg$ of ice at $-{10}^{\circ }C$to increase its temperature to $0^{\circ }C$.
$Q_2=m_i{S}_i\Delta T_i$
where $m_i$ is the mass of the ice, $s_i$ is the specific heat of ice and $\Delta T_i$ is the change in temperature of ice.
Given, $m_i=0.10kg$, $s_i=0.50kcalk{g}^{-1}$
$Q_2=0.10\times (0.5\times {10}^3)\times [0-(-10\left)\right]$
$=500cal$
So, remaining heat,
$Q=Q_1-Q_2=3000-500=2500cal$
Now as latent heat of ice is $l=80kcalk{g}^{-1}$, the remaining heat will melt ice only
$L=\frac{Q}{m}\Rightarrow m=\frac{Q}{m}=\frac{2500}{80}=31.25g$ of ice.
Initial amount of ice $=0.10kg=100g$
So, the remaining ice,
$=100-31.25=68.75g$
Initial amount of water $=0.15kg=150g$
Therefore, total water $=150+31.25=181.25g$
The temperature of the system will be $0^{\circ }C$.