A potential difference of \[100\,{\text{V}}\] is applied across a resistor of resistance \[50\,\Omega \] for 6 minutes and 58 seconds. Find the heat produced in 1) joule 2) calorie.

Hint: Use Ohm’s law to determine the current. Then use the equation of heat produced in terms of current, resistance and time. Convert the unit of time in the SI system of units to determine the heat produced in joule. Use the conversion factor between joule and calorie to determine the heat produced in calorie.

Complete step by step solution:
The expression for Ohm’s law is
\[I = \dfrac{V}{R}\] …… (1)
Here, \[I\] is the current, \[V\] is the potential difference and \[R\] is the resistance.
The heat \[H\] produced is given by
\[H = {I^2}Rt\] …… (2)
Here, \[I\] is the current, \[R\] is the resistance and \[t\] is the time.
(1) Now, It is given that,
\[V = 100\,{\text{V}}\]
\[R = 50\,\Omega \]
\[t = 6\,\min 58\sec \]
Convert the unit of time \[t\] in the SI system of units.
\[t = \left( {6\,\min 58\sec } \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)\]
\[ \Rightarrow t = 418\,\sec \]
Hence, the time in the SI system of units is \[418\,\sec \].
Determine the current \[I\] through the circuit.
Substitute \[50\,\Omega \] for \[R\] and \[100\,{\text{V}}\] for \[V\] in equation (1).
\[I = \dfrac{{100\,{\text{V}}}}{{50\,\Omega }}\]
\[ \Rightarrow I = 2\,{\text{A}}\]
Hence, the current thought the circuit is \[2\,{\text{A}}\].
Determine the heat produced.
Substitute \[2\,{\text{A}}\] for \[I\], \[50\,\Omega \] for \[R\] and \[418\,\sec \] for \[t\] in equation (2).
\[H = {\left( {2\,{\text{A}}} \right)^2}\left( {50\,\Omega } \right)\left( {418\,\sec } \right)\]
\[ \Rightarrow H = 83600\,{\text{J}}\]

Hence, the heat produced in joule is \[83600\,{\text{J}}\].

(2)
One calorie of heat is equal to the heat \[4.184\,{\text{J}}\].
\[1\,{\text{cal}} = 4.184\,{\text{J}}\]
\[ \Rightarrow 1\,{\text{J}} = \dfrac{1}{{4.184}}\,{\text{cal}}\]
Determine the heat produced \[83600\,{\text{J}}\] in calorie.
\[83600\,{\text{J}} = \dfrac{{83600\,{\text{J}}}}{{4.184}}\,{\text{cal}}\]
\[ \Rightarrow 83600\,{\text{J}} = 19980.87\,{\text{cal}}\]

Hence, the heat produced in calorie is \[19980.87\,{\text{cal}}\].

Note: One may also solve the given example in another way. One can determine the power using the relation between the potential difference and resistance and then determine the energy by tasking the product of power and time and then use the conversion factor between joule and calorie. Don’t forget to convert the unit of time in seconds.