A gas is contained in a large cylinder. As you add 325,000 J of heat to gas ther
ID: 1640387 • Letter: A
Question
Explanation / Answer
Let the constant pressure be P = 2,40,000 Pa.
Initial volume be V1 = 0.83 m3
Final volume V2 = 0.45 m3
Then the work done dW = P × ( V2 - V1 )
= 2,40,000 × ( 0.45 - 0.83 )
= - 2,40,000 × 0.38
= - 91,200 J
Negative sign indicates that work is done on gas.
Here heat energy added dQ = 3,25,000 J
From first law of thermodynamics ,
Heat energy dQ = dU + dW
So change in internal energy dU = dQ - dW
Here dW is negative, then dU = dQ + dW
= 3,25,000 + 91,200
= 4,17,200 J
Here since heat energy is added, it is positive.So Internal energy of the gas has increased.
B) Let the specific heat of water be C = 1 cal/g
Now volume of water be V = 227 L = 227 kg
So mass of the water M = 2,27,000 g
We know that 1 L water has a weight 1 kg
Initial temperature of water be T1 = 10
Final temperature of water be T2 = 51
Change in temperature be T = 51 - 10 = 41
Now heat energy absorbed Q = M × C × T
= 2,27,000 × 1 × 41
= 93,07,000 J
Now power of the heater P = 9,400 W = 9400 J/s
So we know that power P = energy / time .
Then Time = energy / power
So time taken to heat 227 L water is T = Q /P
T = 93,07,000 / 9400
= 990.106 sec
Time taken in minutes T = 990.106 / 60 = 16.5 minutes.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.