Question

$1kg$ 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 $540cal.g^{-1}$. specific heat of the steam is $1cal\cdot g^{-1}\cdot {}^{\circ }{C}^{-1}$. 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,
$m_s=1kg=1000g=10^3$
Temperature of steam,
$T_1=150^{\circ }C$
Latent heat of steam,
$L_s=540cal{g}^{-1}$
$c=1cal{g}^{-10}{C}^{-1}$
Heat lost by steam,
$Q^{\prime }=m{L}_f+m_{s}c\Delta t$
$=10^3\times 540+10^3\times 1\times (150-90)$
$=54\times 10^4+6\times 10^4$
$=10^4(54+6)=60\times 10^4$ cal
Heat gained by $20L$ water,
$Q^{\prime \prime }=m_w\times c\times \Delta T$
Mass of $20L$ water, $m_w=20kg=20\times 10^{3}g$
$\therefore Q^{\prime \prime }=20\times 10^3\times 1\times \Delta T$
By the principle calorimetry,
Heat lost $=$ Heat gained
$60\times 10^4=20\times 10^3\times \Delta T$
$\Rightarrow \Delta T=\frac{60\times 10^4}{20\times 10^3}$
$=30^{\circ }C$