# A heating coil is passed in a calorimeter of heat capacity $50\,{\text{J}} \cdot ^\circ {{\text{C}}^{ - 1}}$ containing $1\,{\text{kg}}$ of a liquid of specific heat capacity $450\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{{\text{ - 1}}}} \cdot ^\circ {{\text{C}}^{ - 1}}$. The temperature of liquid rises by $10^\circ {\text{C}}$ when $2.0\,{\text{A}}$ current passes for 10 minutes. Find the resistance of the coil.

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Hint:Use the formula for power in terms of the current and resistance. Also use the formula for power in terms of the change in heat and time interval. Use the formula for the heat exchanged for a substance. This formula gives the relation between mass of the substance, specific heat of the substance and change in temperature of the substance.

Formulae used:
The power $P$ is given by
$P = {I^2}R$ …… (1)
Here, $I$ is the current and $R$ is the resistance.
The power $P$ is given by
$P = \dfrac{{\Delta Q}}{{\Delta t}}$ …… (2)
Here, $\Delta Q$ is the change in the heat in the time interval $\Delta t$.
The heat exchanged $\Delta Q$ is given by
$\Delta Q = mc\Delta T$ …… (3)
Here, $m$ is the mass of the substance, $c$ is specific heat of the substance and $\Delta T$ is a change in temperature of the substance.

We have given that the heat capacity of the calorimeter is $50\,{\text{J}} \cdot ^\circ {{\text{C}}^{ - 1}}$ and the specific heat capacity of the liquid is $450\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{{\text{ - 1}}}} \cdot ^\circ {{\text{C}}^{ - 1}}$.
$c = 50\,{\text{J}} \cdot ^\circ {{\text{C}}^{ - 1}}$
$\Rightarrow{c_L} = 450\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{{\text{ - 1}}}} \cdot ^\circ {{\text{C}}^{ - 1}}$
The mass of the liquid in the calorimeter is $1\,{\text{kg}}$.
$m = 1\,{\text{kg}}$
The change in the temperature of the liquid is $10^\circ {\text{C}}$ and the current passed through the calorimeter is $2.0\,{\text{A}}$ for 10 minutes.
$\Delta T = 10^\circ {\text{C}}$
$\Rightarrow I = 2.0\,{\text{A}}$
$\Rightarrow \Delta t = 10\,{\text{min}}$

Convert the unit of the time to the SI system of units.
$\Delta t = \left( {10\,{\text{min}}} \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)$
$\Rightarrow \Delta t = 600\,{\text{s}}$
We have asked to calculate the resistance of the coil.
Substitute ${I^2}R$ for $P$ in equation (2).
${I^2}R = \dfrac{{\Delta Q}}{{\Delta t}}$
Rearrange the above equation for $R$.
$R = \dfrac{{\Delta Q}}{{{I^2}\Delta t}}$
Substitute $m{c_{net}}\Delta T$ for $\Delta Q$ in the above equation.
$R = \dfrac{{m{c_{net}}\Delta T}}{{{I^2}\Delta t}}$

The net specific heat capacity of the system is the sum of the specific heat capacity of the calorimeter and the liquid in the calorimeter.
${c_{net}} = c + {c_L}$
Substitute $c + {c_L}$ for ${c_{net}}$ in the above equation.
$R = \dfrac{{m\left( {c + {c_L}} \right)\Delta T}}{{{I^2}\Delta t}}$
Substitute $1\,{\text{kg}}$ for $m$, $50\,{\text{J}} \cdot ^\circ {{\text{C}}^{ - 1}}$ for $c$, $450\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{{\text{ - 1}}}} \cdot ^\circ {{\text{C}}^{ - 1}}$ for ${c_L}$, $10^\circ {\text{C}}$ for $\Delta T$, $2.0\,{\text{A}}$ for $I$ and $600\,{\text{A}}$ for $\Delta t$ in the above equation.
$R = \dfrac{{\left( {1\,{\text{kg}}} \right)\left[ {\left( {50\,{\text{J}} \cdot ^\circ {{\text{C}}^{ - 1}}} \right) + \left( {450\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{{\text{ - 1}}}} \cdot ^\circ {{\text{C}}^{ - 1}}} \right)} \right]\left( {10^\circ {\text{C}}} \right)}}{{{{\left( {2.0\,{\text{A}}} \right)}^2}\left( {600\,{\text{A}}} \right)}}$
$\Rightarrow R = 2.08\,\Omega$
$\therefore R \approx 2.1\,\Omega$

Hence, the resistance of the coil is $2.1\,\Omega$.

Note: The students should not forget to convert the unit of the time for which the current is passed through the calorimeter in the SI system of units. The students should also not forget to take the sum of the specific heats of the calorimeter and the liquid in the calorimeter. If this net specific heat of the system is not taken then the final answer will be incorrect.