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').

It's have to be 12 or idk

You will hear a hissing sound because of the carbon dioxide escaping the can or bottle