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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.

\[R = \dfrac{{\left( {1\,{\text{kg}}} \right)\left[ {\left( {50\,{\text{J}} \cdot ^\circ {{\text{C}}^{ - 1}}} \right) + \left( {450\,{\text{J}} \cdot {\text{k}}{{\text{g}}^{{\text{ - 1}}}} \cdot ^\circ {{\text{C}}^{ - 1}}} \right)} \right]\left( {10^\circ {\text{C}}} \right)}}{{{{\left( {2.0\,{\text{A}}} \right)}^2}\left( {600\,{\text{A}}} \right)}}\]

\[ \Rightarrow R = 2.08\,\Omega \]

\[ \therefore R \approx 2.1\,\Omega \]

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