Question

A Nichrome heating element that has resistance 28.0? is connected to a battery that has emf 82.0V and internal resistance 1.6? . An aluminum cup with mass 0.130 kg contains 0.200 kg of water. The heating element is placed in the water and the electrical energy dissipated in the resistance of the heating element all goes into the cup and water. The element itself has very small mass. How much time does it take for the temperature of the cup and water to rise from 21.2 ?C to 34.5 ?C? (The change of the resistance of the Nichrome due to its temperature change can be neglected.)

Answer 1

$current = \frac{voltage}{totalresistance}$

$current = \frac{82}{28+1.6}$

$current = \frac{82}{29.6}$

$current = 2.77027$

total heat required for

water = $m_{water} * s_{water} * \Delta T$

where s is specific heat

so

change in temperature = $\Delta T = 34.5-21.2 = 13.3^{0}C$

$s_{water} = 4.179 \frac{J}{gram*C}}$

$m_{water} = 0.2kg = 200g$

heat required = $m_{water} * s_{water} * \Delta T$

heat required = $200*4.179*13.3 J$

heat required = $11116.14J$

total heat required for

Aluminium = $m_{Aluminum} * s_{Aluminum} * \Delta T$

where s is specific heat

so

change in temperature = $\Delta T = 34.5-21.2 = 13.3^{0}C$

$s_{Aluminium} = 0.9 \frac{J}{gram*C}}$

$m_{Aluminium} = 0.13kg = 130g$

heat required = $m_{Aluminium} * s_{Aluminium} * \Delta T$

heat required = $130*0.900*13.3 J$

heat required = $1556.1J$

Total heat = $11116.14 + 1556.1$

Total heat = $12672.24J$

Heat developed by heating element = $i^{2}R* time$

Equate both

$i^{2}R* time = 12672.24$

$2.77027^{2}* 28* time = 12672.24$

$214.88308*time = 12672.24$

$time = 58.9727$$time = 58.9727sec$

Answer 2