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\(1 \mathrm{~kg}\) of steam at \(150^{\circ} \mathrm{C}\) is passed from a steam chamber is to a copper coil immersed in \(20 \mathrm{~L}\) of water. The steam condenses in the coil and is returned to the steam chamber as water at \(90^{\circ} \mathrm{C}\). Latent heat of steam is \(540 \mathrm{cal} \mathrm{g}^{-1}\), specific heat of the steam is \(1 \mathrm{cal}\) and \(\mathrm{g}^{-1{ }^{\circ}} \mathrm{C}^{-1}\). Then, the rise in temperature of water is
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Verified Answer
The correct answer is:
\(30^{\circ} \mathrm{C}\)
Mass of steam,
\(m_s=1 \mathrm{~kg}=1000 \mathrm{~g}=10^3\)
Temperature of steam,
\(T_1=150^{\circ} \mathrm{C}\)
Latent heat of steam,
\(\begin{aligned}
L_s & =540 \mathrm{cal} \mathrm{g}^{-1} \\
c & =1 \mathrm{cal} \mathrm{g}^{-1{ }^{\circ} \mathrm{C}^{-1}}
\end{aligned}\)
Heat lost by steam,
\(\begin{aligned}
Q^{\prime} & =m L_f+m_s c \Delta t \\
& =10^3 \times 540+10^3 \times 1 \times(150-90) \\
& =54 \times 10^4+6 \times 10^4 \\
& =10^4(54+6)=60 \times 10^4 \mathrm{cal}
\end{aligned}\)
Heat gained by \(20 \mathrm{~L}\) water,
\(Q^{\prime \prime}=m_w \times c \times \Delta T\)
Mass of \(20 \mathrm{~L}\) water, \(m_w=20 \mathrm{~kg}=20 \times 10^3 \mathrm{~g}\)
\(\therefore Q^{\prime \prime}=20 \times 10^3 \times 1 \times \Delta T\)
By the principle calorimetry,
Heat lost \(=\) Heat gained
\(60 \times 10^4=20 \times 10^3 \times \Delta T\)
\(\Rightarrow \quad \Delta T=\frac{60 \times 10^4}{20 \times 10^3}=30^{\circ} \mathrm{C}\)
\(m_s=1 \mathrm{~kg}=1000 \mathrm{~g}=10^3\)
Temperature of steam,
\(T_1=150^{\circ} \mathrm{C}\)
Latent heat of steam,
\(\begin{aligned}
L_s & =540 \mathrm{cal} \mathrm{g}^{-1} \\
c & =1 \mathrm{cal} \mathrm{g}^{-1{ }^{\circ} \mathrm{C}^{-1}}
\end{aligned}\)
Heat lost by steam,
\(\begin{aligned}
Q^{\prime} & =m L_f+m_s c \Delta t \\
& =10^3 \times 540+10^3 \times 1 \times(150-90) \\
& =54 \times 10^4+6 \times 10^4 \\
& =10^4(54+6)=60 \times 10^4 \mathrm{cal}
\end{aligned}\)
Heat gained by \(20 \mathrm{~L}\) water,
\(Q^{\prime \prime}=m_w \times c \times \Delta T\)
Mass of \(20 \mathrm{~L}\) water, \(m_w=20 \mathrm{~kg}=20 \times 10^3 \mathrm{~g}\)
\(\therefore Q^{\prime \prime}=20 \times 10^3 \times 1 \times \Delta T\)
By the principle calorimetry,
Heat lost \(=\) Heat gained
\(60 \times 10^4=20 \times 10^3 \times \Delta T\)
\(\Rightarrow \quad \Delta T=\frac{60 \times 10^4}{20 \times 10^3}=30^{\circ} \mathrm{C}\)
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