Question

A small immersion heater is rated at 345W . The specific heat of water is

4186 J/kg?C?.

Part A

Estimate how long it will take to heat a cup of soup (assume this is 250 mL of water) from 18?C to 71?C. Ignore the heat loss to the surrounding environment.

Answer 1

Answer 2

SOLUTION :

Mass of 250 mL water = 250 g = 0.25 kg

Heat required to heat this water to 71ºC from 18ºC

= m s (t2 - t1)

= 0.25 * 4186 * (71 - 18) 

= 55464.5 J 

Let time taken is t sec by the 345 W immersion heater .

Electrical energy consumed

 = 345 * t      watt-sec 

= 345 t        J ( since 1 watt-sec = 1 J)

As per conservation of energy :

Electrical energy consumed = Heat energy generated to heat the water

=> 345 t = 55464.5 

=> t = 55464.5 / 345 = 160.77 sec = 2 min. 40.77 secs.

Time needed to heat the water as required = 160.77 secs = 2 minutes 40.77 secs (ANSWER).