Question

1. If a closed system loses 2200 J of heat (to the surrounding environment) and 800 J of work is done on the system then what was the change in the internal energy of the system.

2. An ideal gas (inside a cylinder fitted with a movable piston) is heated and slowly expands at constant pressure of 5 atm from 1.5 L to 5.5 L.  How much work is done by the gas in this process?

3. If 1,700 J of heat was added to the gas in problem 3, then how much did the internal energy of the system change?

4. What is the efficiency of an automobile engine that has a heat input of 60.0 kJ and a heat output of 25.0 kJ?

5. If the exhaust temperature of the engine in problem 5 is 250° C, then what is the intake temperature?    (The exhaust and intake temperatures are the low and high operating temperatures of the engine, T_L and T_H)

Answer 1

If a closed system loses 2200 J of heat (to the surrounding environment) and 800 J of work is done on the system then what was the change in the internal energy of the system.

Please notice the sign I used  

ΔQ = W+ ΔU

-2200 = -800 +ΔU

ΔU = -2200+800 = -1400 J answer

An ideal gas (inside a cylinder fitted with a movable piston) is heated and slowly expands at constant pressure of 5 atm from 1.5 L to 5.5 L.  How much work is done by the gas in this process?

work done if pressure is constant

P = 5 atm = 5*101325 = 506625 pascals

Vi =1.5L = 1.5*10^-3 m³

Vf = 5.5 L = 5.5*10^-3 m³

W =P*ΔV = 506625 *(  5.5*10^-3 - 1.5*10^-3 ) =2026.5 J answer

If 1,700 J of heat was added to the gas in problem 3, then how much did the internal energy of the system change?

ΔQ = W+ ΔU

ΔU = ΔQ-W = +1700-2026.5 = -326.5 J

What is the efficiency of an automobile engine that has a heat input of 60.0 kJ and a heat output of 25.0 kJ?If the exhaust temperature of the engine in problem 5 is 250° C, then what is the intake temperature?    (The exhaust and intake temperatures are the low and high operating temperatures of the engine

η = 1 - T_L/T_H = 1- Q_L /Q_H = work /heat input

= ( 60-25 ) /60 = 0.583333333
   efficiency =0.583333333

or 58.333 %

1 - T_L/T_H =0.583333333

T_L/T_H = 1-0.583333333 = 0.416666667

T_L/T_H = 0.416666667

T_H = T_L /0.416666667 = ( 250+273)/0.416666667 = 1255.2

intake temperature =1255.2 K

or =1255.2-273 = 982.2° C

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