Question

A small electric immersion heater is used to heat 57 g of water for a cup of instant coffee. The heater is labeled "140 watts" (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from 22°C to 100°C, ignoring any heat losses. (The specific heat of water is 4186 J/kg-K.) Number Units the tolerance is +/-2%

Answer 1

Solution :-

Energy needed to raise temperature,
q = m*c*ΔT
q = 57* 4.186 * (100 - 22) J
q = 18610.956 J

Energy = Power * time
18610.956 = 140 * t
t = 132.93 sec

Time required, t = 132.93 s or 2.215 minutes

Answer 2

SOLUTION :

Mass of water = 57 g = 0.057 kg

Heat required to heat this water to 100ºC from 22ºC

= m s (t2 - t1)

= 0.057* 4186 * (100 - 22) 

= 18610.96 J 

Let time taken is t sec by the 140 W immersion heater .

Electrical energy consumed

 = 140 * t      watt-sec 

= 140 t        J ( since 1 watt-sec = 1 J)

As per conservation of energy :

Electrical energy consumed = Heat energy generated to heat the water

=> 140 t = 18610.96 

=> t = 18610.96 / 140 = 132.94 sec = 2 min. 12.94 secs.

=> Units consumed = 18610.96 / (1000 * 3600) = 0.0052 units .

(1 KWh = 1 unit)

Time needed to heat the water as required = 132.94 secs = 2 minutes 12.94 secs .

Units consumed = 0.0052 .

(ANSWER).