Practice question
Question
How much heat is required to raise the temperature of 0.3 kg of lead from 35^circ C to 65^circ C ? (Specific heat of lead = 127.7 J kg⁻¹ K⁻¹ )
Explanation
**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h <1, impossible 100% efficient, Carnot efficiency η_Carnot =1 - T_c/T_h maximum possible between T_h and T_c. Δ Q = m s Δ T . m = 0.3 , s = 127.7 , Δ T = 65 - 35 = 30 . Δ Q = 0.3 × 127.7 × 30 = 1149.3 J ≈ 1149 J . Using first law ΔU = Q - W, W =
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