Joules law of heating

Electricity and Magnetism Level 2

A heating coil immersed in a calorimeter of heat capacity 50 J/C containing 1 kg of a liquid of specific heat capacity 450 J/Kg/C. The temperature of liquid rises by 10 degrees when a 2A current is passed for 10 minutes. Find the potential difference across the coil. Assume that there is no loss of heat.

The answer is 4.16.

1 solution

Tom Engelsman
Sep 13, 2017

The required formulation is:

$P = \frac{(c_{coil} + c_{liquid})\Delta T}{t} = VI$

and solving for the potential difference (voltage) gives:

$V = \frac{(c_{coil} + c_{liquid})\Delta T}{It}$ .

Plugging in the required values yields:

$V = [(50 J/C + (1kg)(450 J/kgC))(10 C) / (2 A)(600 sec)] \cdot \frac{1 V}{1 J/A*sec} = \boxed{4.167 V}.$

Joules law of heating

The answer is 4.16.

1 solution

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