An electric heater supplies heat at the rate of 1000J per second. How much time is required to raise the temperature of 200 g of water from 20°C to 90°C? (58.8 s)
Difficulty: Medium
Solution: Power = P = 1000 $Js^{-1}$
Mass of water = m = 200 g =$\frac{200}{1000}$ = 0.2 kg
Initial temperature = $T_{2}$= 20°C = 20 + 273 = 293 K
Final temperature = $T_{1}$= 90°C = 90 + 273 = 363 K
Change in temperature = $\triangle T$ = $T_{2} - T_{1}$ = 363 – 293 = 70 K
Specific heat of water = c = $4200 J〖kg〗^{-1} K^{-1}$
Time = t = ?
P =$\frac{w}{t}$
Or P =$\frac{Q}{t}$
Or P $\times t$ = Q
Or P $\times t$ = $mc \triangle T$
Or t=$\frac{(mc \triangle T)}{P}$
t = $\frac{(0.2\times 4200 \times 70)}{1000}$= 58.8 s
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