Question

Ice at $-{20}^{\circ }C$ is added to $50g$ of water at ${40}^{\circ }C$. When the temperature of the mixture reaches $0^{\circ }C$, it is found that $20g$ of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water $=4.2J/g^{\circ }C$) Specific heat of Ice $=2.1J/g^{\circ }C$, Heat of fusion of water at $0^{\circ }C=334J/g$

• A. $40g$
• B. $50g$
• C. $100g$
• D. $60g$

Answer 1

The correct option is A $40g$

We know that: $Q=mL,Q=ms\Delta \theta$

Given: ${\theta }_{ice}=-{20}^{\circ }C,m_w=50g,{\theta }_w={40}^{\circ }C$,
$m_{remain}=20g,s_w=4.2J/g^{\circ }C,s_{ice}=2.1J/g^{\circ }C,L=334J/g$

Let amount of ice is $m$ gram.
According to principle of calorimeter heat taken by ice is heat given by water
$m{s}_{ice}(0-(-20\left)\right)+(m-m_{remian})L=m_w{s}_w(40-0)$
$20\times 2.1\times m+(m-20)\times 334=50\times 4.2\times 40$
$376m=8400+6680$
$m=40.1g$
$m\approx 40g$

Final Answer: (b)