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In an experiment on the specific heat of a metal, a $0.20$ kg block of a metal at $150^{\circ} \mathrm{C}$ is dropped in a copper calorimeter (of water equivalent $0.025 \mathrm{~kg}$ ) containing $150 \mathrm{~cm}^3$ of water at $27^{\circ} \mathrm{C}$. The final temperature is $40^{\circ} \mathrm{C}$. Compute the specific heat of the metal . If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
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Verified Answer
Mass of the metal $m_1=0.2 \mathrm{~kg}=200 \mathrm{~g}$
Fall in temperature $=\Delta t_1=150-40=110^{\circ} \mathrm{C}$
Heat lost by the metal $=\Delta Q_1=\operatorname{mc} \Delta t_1$
$[\mathrm{c}=$ specific heat of metal]
$$
=200 \times c \times 110
$$
Volume of water $=150 \mathrm{~cm}^3 ;$ Mass of water $=m_2=150 \mathrm{~g}$
Water equivalent of calorimeter
$$
=w=0.025 \mathrm{~kg}=25 \mathrm{~g}
$$
Rise in temperature of water and calorimeter
$$
\Delta t_2=40-27=13^{\circ} \mathrm{C}
$$
Heat gained by water and calorimeter
$$
\begin{aligned}
&=\Delta Q_2=\left(m_2+w\right) \Delta t_2=(150+25) \times 13 \\
&=175 \times 13 \\
&\because \quad \text { Heat lost }=\text { heat gained, } \\
&\quad 200 \times c \times 110=175 \times 13 \\
&\quad c=\frac{175 \times 13}{200 \times 110}=0.1
\end{aligned}
$$
The value of $\mathrm{c}$ is less than its actual value, as some heat is lost to the surroundings.
Fall in temperature $=\Delta t_1=150-40=110^{\circ} \mathrm{C}$
Heat lost by the metal $=\Delta Q_1=\operatorname{mc} \Delta t_1$
$[\mathrm{c}=$ specific heat of metal]
$$
=200 \times c \times 110
$$
Volume of water $=150 \mathrm{~cm}^3 ;$ Mass of water $=m_2=150 \mathrm{~g}$
Water equivalent of calorimeter
$$
=w=0.025 \mathrm{~kg}=25 \mathrm{~g}
$$
Rise in temperature of water and calorimeter
$$
\Delta t_2=40-27=13^{\circ} \mathrm{C}
$$
Heat gained by water and calorimeter
$$
\begin{aligned}
&=\Delta Q_2=\left(m_2+w\right) \Delta t_2=(150+25) \times 13 \\
&=175 \times 13 \\
&\because \quad \text { Heat lost }=\text { heat gained, } \\
&\quad 200 \times c \times 110=175 \times 13 \\
&\quad c=\frac{175 \times 13}{200 \times 110}=0.1
\end{aligned}
$$
The value of $\mathrm{c}$ is less than its actual value, as some heat is lost to the surroundings.
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