An electric heater supplies heat at the rate of 1000 joules per second. How much time is required to raise the temperature of 200g of water from 20°C to 90°C? Ans. (58.8 s)


Given:

Rate of heat Transferred = R = 100 J s⁻¹ 
Mass of water  = m = 200 g = 0.2 Kg
Initial Temperature = T₁ = 20°C = (20 + 273) K = 293 K
Final Temperature = T₂ = 90°C = (90 + 273) K = 363 K
Change in Temperature = ΔT = T₂  - T₁ = 363 K - 293 K = 70 K
Specific heat of water  = c = 4200 J Kg⁻¹ K⁻¹

To Find:

Time required  = t  = ? sec

Solution:  

Time can be calculated from the heat transfer formula which is 

R = `\frac {Q}{t}`

or

t = `\frac {Q}{R}`   ------eqn (1)

So we have to find first the Heat Q So, 

The formula for heat capacity is 

Q  = mcΔT

By Putting values

Q  = 0.2 Kg ｘ 4200 J Kg⁻¹ K⁻¹ ｘ 70 K

Q  = 58,800 J

Now using Eqn (1)

t = `\frac {Q}{R}` 

by putting values

t = `\frac {58,800 J}{100 J s⁻¹ }`

t = 588 s

thus the time will consume 588 seconds

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