An electric heater supplies heat at the rate of 1000 joules per second . How much time is required to raise the temperature of 200 g of water from 20 C to 90 C .
Answers
Answer:
Rate of heat = 1000 joules / second
Time required to raise the temperature = 1000/60
=16.66 °C
Time required to raise temperature of 200g = 200/60
= 3.33 °C
Water at 20 to 90° C = 90-20
=70°C
= 70*3.33
233.333 °C
Explanation:
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Given Data
Mass of water=200g = 200/1000 = 0.2kg
Initial temperature= T₁ = 20 C =273+20=293K
Final temperature= T₂ = 90 C = 273+90=363K
Power of heat= P =1000J/s
Required
Time = t = ?
Formula
Power = W/t =Q/t
t = Q/P
Q = mc(T₂ - T₁)
Calculation
Q = mc(T₂ - T₁)
= 0.2 × 4200 × (363 - 293)
= 58800J
Time= t = Q/P
= 58800/1000
= 58.8s
Answer
Time = t = 58.8s