Problem Statement
Solve the thermodynamics problem: Mars is 1.52 AU from the Sun (Earth is 1 AU). If Earth’s effective temperature is 280 K, find Mars’s effective temperature. (Assume both planets are perfect black bodies.) All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section)
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Thermodynamics governs energy transformations involving heat and work. The First Law $\Delta U = Q – W$ expresses energy conservation. For an ideal gas, internal energy depends only on temperature ($U = nC_VT$), and the equation of state $PV = nRT$ links pressure, volume, and temperature.
- $\Delta U = Q – W$ — First Law of Thermodynamics
- $PV = nRT$ — ideal gas equation
- $C_P – C_V = R$, $\gamma = C_P/C_V$
- $W = \int P\,dV$ — work done by gas
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\eta = 1 – \frac{300}{600} = 1 – 0.5 = 0.50 = 50\%$$
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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