A 1.0 kw heater supplies energy to a liquid of mass 1 kg. The temperature of the liquid changes by 80 Kin
a time of 400 s. The specific heat capacity of the liquid is 4.0 kJ/kg*K. What is the average power lost by
the liquid?
100
200
400
800
Answers
Answer:
200w
Explanation:
power consumption by liquid is
Q = m x s x ∆ T
= 1 x 4000x 80
= 3.2 x 10⁵j
power used = Q /t = 3.2 x 10⁵ / 400
= 8 x 10² = 800W
average power lost by liquid = 1000 - 800 = 200W
Given info : A 1 kW heater supplies energy to a liquid of mass 1 kg . The temperature of the liquid changes by 80K in a times of 400 sec. the specific heat of capacity of the liquid is 4.0 kJ/kg/K.
To find : the average power lost by the liquid is..
solution : output energy = power of heater × time
= 1 kW × 400s
= 400 kJ
input energy = energy needed to change the temperature of liquid = msΔT
= 1 kg × 4 kJ/kg/K × 80K
= 320 kJ
therefore the lost in energy by the liquid = output energy - input energy
= 400 kJ - 320 kJ
= 80 kJ
now the average power lost by the liquid = the lost in the energy by the liquid/time = 80kJ/400 = 80000J/400s = 200 W
therefore the correct option is (b) 200