The heating coil of a hot water heater has a resistance of 17 Ω and operates at 300 V. How long a time is required to raise the temperature of 115 kg of water from 18°C to 75°C? (The specific heat for water = 10 3 cal/kg ⋅°C and 1.0 cal = 4.186 J).

5183 sec.

Explanation:

The power dissipated in the resistance, will be used to raise the temperature of the water completely.This power can be found applying Joule's law, which states the following:

       

If the resistance of the heating coil can be assumed as constant with the temperature, we can find the current I applying Ohm's Law, as follows:

       

Now, from (1) we can find P, as follows:

       

The energy supplied by this power, is just the product of the power times the time during which the energy was delivered:

       

This energy, will be supplied as heat to the mass of water, as stated by the following equation (assuming no heat losses out of the heater):

       

where c= specific heat of water = 4186 J/ºC*kg, m= 115 kg, and ΔT= 57ºC.From (2) and (3), if left sides are equal each other, so do right sides:

       

Replacing by the values, we can solve for  Δt, as follows:

       

The time required to raise the temperature of  115 kg of  water, from 18ºC, to 75ºC, is 5183 sec (approximately 1hr 26').