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. What is the final mass of water in the vessel? (Specific heat of ice $=0.5{\text{cal g}}^{-1}$${}^{\circ }{C}^{-1}$).

• A. $700\text{g}$
• B. $600\text{g}$
• C. $400\text{g}$
• D. $200\text{g}$

Answer 1

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

Given that,
Mass of ice, $m_i=200\text{g}$
Temperature of ice, $T_i=-{20}^{\circ }\text{C}$
Mass of water, $m_w=500\text{g}$
Temperature of water, $T_w={20}^{\circ }\text{C}$
Specific heat capacity of ice, $s_i=0.5{\text{cal g}}^{-1}$${}^{\circ }{\text{C}}^{-1}$
Specific heat capacity of water, $s_w=1{\text{cal g}}^{-1}$${}^{\circ }{\text{C}}^{-1}$

Maximum heat supplied by water is,
$Q=m_w{s}_w\Delta T$
$\Delta Q_1=500\times 1\times (20-0)$
$\Rightarrow \Delta Q_1=10000\text{cal}$
Heat required to raise the temperature of ice up to $0^{\circ }\text{C}$ is,
$\Delta Q_2=m_i{s}_i\Delta T$
$\Rightarrow \Delta Q_2=\left(200\right)\left(0.5\right)\left(20\right)$
$\therefore \Delta Q_2=2000\text{cal}$
Heat required to melt ice is
$\Delta Q_1-\Delta Q_2=10000-2000=8000\text{cal}$
$\Rightarrow \Delta Q_1-\Delta Q_2=m{L}_f$
$\therefore m=\frac{8000}{80}=100\text{g}$
$(\therefore L_f=80\text{cal/g})$
Total mass of water in vessel is,
$\Rightarrow m_{final}=500+100=600\text{g}$