The electrical energy E {electrical supplied by the heater goes into increasing the temperature of the aluminium cup and the} water a temperature difference\({\rm{\Delta T}}\):
\({{\rm{E}}_{{\rm{electrical }}}} = {{\rm{Q}}_{{\rm{Al}}}}{\rm{ + }}{{\rm{Q}}_{{\rm{water }}}}\)
Here,\({{\rm{Q}}_{{\rm{Al}}}}\)is the energy required to raise the temperature of the aluminium cup and\({{\rm{Q}}_{{\rm{water }}}}\)is the energy required to raise the temperature of the water.
Here,\({{\rm{E}}_{{\rm{electrical }}}}\)is found in terms of the power output of the heater and the time interval of the heating process by equation (1)\({{\rm{Q}}_{{\rm{Al}}}}\)and\({{\rm{Q}}_{{\rm{water }}}}\)are found from Equation (2):
\(\begin{aligned}{}{\rm{P\Delta t}} &= {{\rm{m}}_{{\rm{Al}}}}{{\rm{c}}_{{\rm{Al}}}}{\rm{\Delta T + }}{{\rm{m}}_{{\rm{water }}}}{{\rm{c}}_{{\rm{water }}}}{\rm{\Delta TP\Delta t}}\\ &= \left( {{{\rm{m}}_{{\rm{Al}}}}{{\rm{c}}_{{\rm{Al}}}}{\rm{ + }}{{\rm{m}}_{{\rm{water }}}}{{\rm{c}}_{{\rm{water }}}}} \right){\rm{\Delta T}}\end{aligned}\)
Substitute for P from Equation (4):
\(\frac{{{{{\rm{(\Delta V)}}}^{\rm{2}}}}}{{\rm{R}}}{\rm{\Delta t = }}\left( {{{\rm{m}}_{{\rm{Al}}}}{{\rm{c}}_{{\rm{Al}}}}{\rm{ + }}{{\rm{m}}_{{\rm{water }}}}{{\rm{c}}_{{\rm{water }}}}} \right){\rm{\Delta T}}\)
Solve the above equation for R as below.
\(R = \frac{{{{(\Delta V)}^2}\Delta t}}{{\left( {{m_{Al}}{c_{Al}} + {m_{water{\rm{ }}}}{c_{water{\rm{ }}}}} \right)\Delta T}}\)
Substitute numerical values in the above equation.
\(\begin{aligned}{}R &= \frac{{{{\left( {120{\rm{ }}V} \right)}^2} \times 120{\rm{ }}s}}{{\left( {\left( {0.100{\rm{ }}\;kg} \right)\left( {900{\rm{ }}{J \mathord{\left/ {\vphantom {J {kg^\circ C}}} \right.\\} {kg^\circ C}}} \right) + \left( {0.350{\rm{ }}\;kg} \right)\left( {4186{\rm{ }}{J \mathord{\left/ {\vphantom {J {kg^\circ C}}} \right.\\} {kg^\circ C}}} \right)} \right)\left( {75.0^\circ C} \right)}}\\ &= 14.8{\rm{ }}\Omega \end{aligned}\)
Therefore, resistance of immersion heater for given temperature is \(14.8{\rm{ }}\Omega \).