Question

$1\, kg$ of steam at $150^{\circ} C$ is passed from a steam chamber in to a copper coil immersed in $20$ litres of water. The steam condenses in the coil and is returned to the steam chamber water at $90^{\circ} C$. Latent heat of steam is $540\, cal.\,g^{-1}$. specific heat of the steam is $1\, cal \cdot g^{-1} \cdot{ }^{\circ} C ^{-1}$. Then, the rise in temperature of water is

• A. $75^{\circ} C$
• B. $60^{\circ} C$
• C. $30^{\circ} C$
• D. $20^{\circ} C$

Answer 1

Answer: (C)

Mass of steam,
$m_{s}=1\, kg =1000\, g =10^{3}$
Temperature of steam,
$T_{1}=150^{\circ} C$
Latent heat of steam,
$L_{s}=540\,cal\, g^{-1}$
$c=1\, cal\, g^{-10} C ^{-1}$
Heat lost by steam,
$Q'=m L_{f}+m_{s} c \Delta t$
$=10^{3} \times 540+10^{3} \times 1 \times(150-90)$
$=54 \times 10^{4}+6 \times 10^{4}$
$=10^{4}(54+6)=60 \times 10^{4}$ cal
Heat gained by $20\, L$ water,
$Q''=m_{w} \times c \times \Delta T$
Mass of $20\, L$ water, $m_{w}=20\, kg =20 \times 10^{3} g$
$\therefore Q''=20 \times 10^{3} \times 1 \times \Delta T$
By the principle calorimetry,
Heat lost $=$ Heat gained
$60 \times 10^{4}=20 \times 10^{3} \times \Delta T$
$\Rightarrow \Delta T=\frac{60 \times 10^{4}}{20 \times 10^{3}}$
$=30^{\circ} C$