- Home
- Standard 11
- Physics
When $0.15\; kg$ of $1 ce$ at $0^{\circ} C$ is mixed with $0.30 \;kg$ of water at $50^{\circ} C$ in a container, the resulting temperature is $6.7^{\circ} C$. Calculate the heat of fuston of ice. $(s_{\text {water }}=4186 J kg ^{-1} K ^{-1}$ ).
$6.54 \times 10^{3} \;J kg ^{-1}$
$6.48 \times 10^{3} \;kJ kg ^{-1}$
$3.34 \times 10^{5} \;kJ kg ^{-1}$
$3.34 \times 10^{5} \;J kg ^{-1}$
Solution
Heat lost by water $=m s_{w}\left(\theta_{{f}}-\theta_{1}\right)_{w}$
$=(0.30 kg )\left(4186 J kg ^{-1} K ^{-1}\right)\left(50.0^{\circ} C -6.7^{\circ} C \right)$
$=54376.14 J$
Heat required to melt $1 ce =m_{2} L_{ f }=(0.15 kg ) L_{ r }$
Heat required to raise temperature of $1 ce$ water to final temperature $=m_{1} s_{w}\left(\theta_{1}-\theta_{1}\right)_{1}$ $=(0.15 kg )\left(4186 J kg ^{-1} K ^{-1}\right)\left(6.7^{\circ} C -0^{\circ} C \right)$
$=4206.93 J$
Heat lost $=$ heat gained $54376.14 J =(0.15 kg ) L_{ f }+4206.93 J$
$L_{1}=3.34 \times 10^{5} \;J kg ^{-1}$
