Question

Ice in a freezer is at $-7^{\circ }C.100g$ of this ice is mixed with $200g$ of water at ${15}^{\circ }C$. Take the freezing temperature of water to be $0^{\circ }C$, the specific heat of ice equal to $2.2J/g^{\circ }C$, specific heat of water equal to $4.2J/g^{\circ }C$, and the latent heat of ice equal to $335J/g$. Assuming no loss of heat to the environment, the mass of ice in the final mixture is closest to

• A. $88g$
• B. $67g$
• C. $54g$
• D. $45g$

Answer 1

The correct option is B $67g$

We will solve this problem by converting the given system at $o^{o}C$ i.e. $100g$ ice at $0^{o}C$ and $200g$ water at $0^{o}C$ and then supply the liberated energy.
Heat required in converting ice from $-7^{o}C$ to $0^{o}C$,
$H_1=m_{ice}{S}_{ice}(T_f-T_i)=100\times 2.2\times 7=1540J$
Heat released in converting water from ${15}^{o}C$ to $0^{o}C$,
$H_2=m_{water}{S}_{water}(T_i-T_f)=200\times 4.2\times 15=12600J$
Net heat released $H_r=12600-1540=11060J$
Now out system is $100g$ ice at $0^{o}C$ and $200g$ water at $0^{o}C$ and a heat energy of $11060J$ is given to it.
So the ice starts to melt at first.
Mass of ice melted $m_{ice}^{\prime }=\frac{11060}{L_{ice}}=\frac{11060}{335}=33g$
Thus ice remained $M_{ice}=100-33=67g$