Physics: Heat and Thermodynamics
Energy transfer, thermal processes, and the laws of thermodynamics
Physics: Heat and Thermodynamics
Energy transfer, thermal processes, and the laws of thermodynamics
Physics - Grade advanced
- 1
A 0.250 kg block of copper at 95.0°C is placed into 0.400 kg of water at 22.0°C in an insulated container. The specific heat of copper is 385 J/kg·°C and the specific heat of water is 4186 J/kg·°C. Assume no heat is lost to the container. Find the final equilibrium temperature.
Set heat lost by the hot object equal to heat gained by the cold object.
The heat lost by the copper equals the heat gained by the water, so 0.250(385)(95.0 - T) = 0.400(4186)(T - 22.0). Solving gives T = 25.98°C, so the final equilibrium temperature is about 26.0°C. - 2
An ideal monatomic gas expands isothermally at 300 K from 2.00 L to 8.00 L. The gas contains 0.500 mol. Calculate the work done by the gas.
For an isothermal ideal gas process, temperature is constant and ΔU = 0.
For an isothermal ideal gas expansion, W = nRT ln(Vf/Vi). Substituting gives W = (0.500)(8.314)(300) ln(8.00/2.00) = 1.73 × 10^3 J. The work done by the gas is about 1.73 kJ. - 3
A gas undergoes a process in which 750 J of heat is added to the gas and 420 J of work is done by the gas. Using the convention ΔU = Q - W, find the change in internal energy.
Using ΔU = Q - W, the change in internal energy is ΔU = 750 J - 420 J = 330 J. The internal energy of the gas increases by 330 J. - 4
A heat engine absorbs 2400 J of heat from a hot reservoir and rejects 1500 J of heat to a cold reservoir during each cycle. Find the work output and the thermal efficiency.
For a cyclic engine, the net change in internal energy over one cycle is zero.
The work output is W = Qh - Qc = 2400 J - 1500 J = 900 J. The efficiency is e = W/Qh = 900/2400 = 0.375, so the engine is 37.5% efficient. - 5
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is its maximum possible efficiency?
The Carnot efficiency is e = 1 - Tc/Th = 1 - 300/600 = 0.500. The maximum possible efficiency is 50.0%. - 6
A 2.00 mol sample of an ideal monatomic gas is heated at constant volume from 250 K to 400 K. Calculate the heat added. Use Cv = (3/2)R.
At constant volume, no boundary work is done by the gas.
At constant volume, Q = nCvΔT. Using Cv = (3/2)R, Q = (2.00)(1.5)(8.314)(400 - 250) = 3.74 × 10^3 J. The heat added is about 3.74 kJ. - 7
A 1.50 mol sample of ideal diatomic gas expands adiabatically from 1.00 L to 3.00 L. Its initial temperature is 500 K. Use γ = 1.40. Find the final temperature.
Use the adiabatic relation involving temperature and volume.
For a reversible adiabatic process, TV^(γ - 1) is constant. Thus Tf = Ti(Vi/Vf)^(γ - 1) = 500(1.00/3.00)^0.40 = 322 K. The final temperature is about 322 K. - 8
A 0.600 kg sample of ice at 0°C is completely melted into water at 0°C. The latent heat of fusion of water is 3.34 × 10^5 J/kg. How much heat is required?
During melting at constant temperature, Q = mLf. Substituting gives Q = (0.600)(3.34 × 10^5) = 2.00 × 10^5 J. The required heat is about 200 kJ. - 9
A copper rod of length 0.750 m and cross-sectional area 1.20 × 10^-4 m^2 connects two thermal reservoirs at 100°C and 20°C. The thermal conductivity of copper is 400 W/m·K. Find the steady-state rate of heat conduction through the rod.
Use the temperature difference in kelvins or degrees Celsius because only the difference matters.
The conduction rate is P = kAΔT/L. Substituting gives P = (400)(1.20 × 10^-4)(80)/(0.750) = 5.12 W. Heat flows through the rod at a rate of 5.12 W. - 10
A blackbody radiator has a surface area of 0.0500 m^2 and temperature 800 K. Its surroundings are at 300 K. Use σ = 5.67 × 10^-8 W/m^2·K^4 and emissivity e = 1.00. Calculate the net radiated power.
The net radiated power is P = eσA(T^4 - Ts^4). Substituting gives P = (1.00)(5.67 × 10^-8)(0.0500)(800^4 - 300^4) = 1.14 × 10^3 W. The net power radiated is about 1.14 kW. - 11
One mole of an ideal gas expands freely into a vacuum from volume V to volume 4V in an insulated container. For an ideal gas, determine Q, W, ΔU, and ΔS.
Free expansion is irreversible, but entropy can be calculated using a reversible path between the same initial and final states.
For free expansion into a vacuum, W = 0 because there is no opposing pressure, and Q = 0 because the container is insulated. For an ideal gas, internal energy depends only on temperature, so ΔU = 0 and the temperature does not change. The entropy change is ΔS = nR ln(Vf/Vi) = R ln 4 = 11.5 J/K. - 12
A refrigerator removes 450 J of heat from its cold interior while 150 J of work is done on it during one cycle. Find the heat rejected to the room and the coefficient of performance.
Energy conservation gives Qh = Qc + W = 450 J + 150 J = 600 J. The coefficient of performance for a refrigerator is COP = Qc/W = 450/150 = 3.00. - 13
An ideal gas follows a rectangular cycle on a P-V diagram with corners at (V, P), (3V, P), (3V, 2P), and (V, 2P), traversed clockwise. Find the net work done by the gas in one cycle in terms of P and V.
On a P-V diagram, the area inside a closed cycle equals the net work.
The net work done by a clockwise cycle equals the area enclosed on the P-V diagram. The rectangle has width 3V - V = 2V and height 2P - P = P, so Wnet = 2PV. The gas does 2PV of work per cycle. - 14
A 3.00 mol sample of ideal gas is compressed isobarically at 2.50 × 10^5 Pa from 0.0400 m^3 to 0.0150 m^3. Find the work done by the gas and state whether energy is transferred into or out of the gas by work.
For an isobaric process, W = PΔV = (2.50 × 10^5)(0.0150 - 0.0400) = -6.25 × 10^3 J. The work done by the gas is negative, meaning 6.25 kJ of energy is transferred into the gas by work done on it. - 15
A 1.00 mol sample of an ideal gas is taken reversibly and isothermally at 350 K from volume V to volume 2V. Find the entropy change of the gas.
Entropy change is a state function, and for an isothermal ideal gas process it depends on the volume ratio.
For a reversible isothermal expansion of an ideal gas, ΔS = nR ln(Vf/Vi). Substituting gives ΔS = (1.00)(8.314) ln 2 = 5.76 J/K. The entropy of the gas increases by 5.76 J/K.