Question

$250g$ of water at ${30}^{\circ }\text{C}$ is contained in a copper vessel of mass $50g$. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to $5^{\circ }\text{C}$. Given specific latent heat of fusion of ice is$336\times {10}^{3}Jk{g}^{-1}$, specific heat capacity of copper is$400Jk{g}^{-1}{K}^{-1}$, and the specific heat capacity of water is $4200Jk{g}^{-1}{K}^{-1}$.

Answer 1

Step 1: Calculate the heat energy given out by the water in reducing its temperature from ${30}^{\circ }\text{C}$ to $5^{\circ }\text{C}$
Mass of water, $m_w=250g=0.25kg$
Initial temperature of water, $T_w={30}^{\circ }\text{C}=303K$
Final temperature of water, $T=5^{\circ }\text{C}=278K$ Specific heat capacity of water,$c_w=4200Jk{g}^{-1}{K}^{-1}$
$\left|\Delta T\right|=|T-T_w|$
$\left|\Delta T\right|=|278K-303K|=25K$
Therefore,
$Q=m_w{c}_w\left|\Delta T\right|$
$\Rightarrow Q_w=\left(0.25\right)\times \left(4200Jk{g}^{-1}{K}^{-1}\right)\times \left(25K\right)$
$\Rightarrow Q_w=26250J$

Step 2: Calculate the heat energy given out by the copper vessal in reducing its temperature from ${30}^{\circ }\text{C}to{5}^{\circ }\text{C}$
Mass of copper vessel, $m_{cu}=50g=0.05kg$
Initial temperature of copper vessel, $T_{cu}={30}^{\circ }\text{C}=303K$
Final temperature of copper vessel, $T=5^{\circ }\text{C}=278K$
Specific heat capacity of water, $c_{cu}=400Jk{g}^{-1}{K}^{-1}$
$|\Delta T|=|T-T_{cu}|$
$|\Delta T|=|278K-303K|=25K$
Therefore,
$Q=m_{cu}{c}_{cu}|\Delta T|$
$\Rightarrow Q_{cu}$
$=\left(0.05kg\right)\times \left(400Jk{g}^{-1}{K}^{-1}\right)\times \left(25K\right)$
$\Rightarrow Q_{cu}=500J$

Step 3: Calculate the heat energy taken by the ice to melt at $0^{\circ }\text{C}$
Mass of ice, $m$
Specific latent heat of fusion of ice, $L=336000Jk{g}^{-1}$
Specific latent heat, $L=\frac{Q}{m}(atT=0^{\circ }\text{C})$
$\Rightarrow Q=mL$
Therefore,
$Q_1=\left(m\right)\times \left(336000Jk{g}^{-1}\right)$
$Q_1=336000mJ$

Step 4: Calculate the heat energy taken by the melted ice to increase its temperature to $5^{\circ }\text{C}$
Mass of water, $m$
Initial temperature of water,$T_{w1}=0^{\circ }\text{C}=273K$
Final temperature of water, $T=5^{\circ }\text{C}=278K$
Specific heat capacity of water, $c=4200Jk{g}^{-1}{K}^{-1}$
$|\Delta T|=|T-T_w|$
$|\Delta T|=|278K-273K|=5K$
Therefore,
$Q=mc|\Delta T|$
$Q_2=m\times 4200\times 5$
$\Rightarrow Q_2=21000mJ$

Step 5: Calculate the mass of ice
Heat energy given out by water and vessel = Heat energy taken by ice
$Q_w+Q_{cu}=Q_1+Q_2$
$\left(26250J\right)+\left(500J\right)=\left(336000mJ\right)+\left(21000mJ\right)$
$\Rightarrow m=\frac{(26250+500)}{336000+21000}$
$\Rightarrow m=74.9g$
The mass of ice required to bring down the temperature of the vessel and its contents to $5^{\circ }\text{C}$ is $74.9g$