Question

The efficiency of a Carnot engine is 30%. Its efficiency is to be raised to 60%. By how much must the temperature of the source be increased if the sink is at 27°C?

Answer 1

Answer on Question #66056, Physics / Mechanics | Relativity

The efficiency of a Carnot engine is 30%. Its efficiency is to be raised to 60%. By how much must the temperature of the source be increased if the sink is at 27°C?

Find: $\Delta T - ?$

Given:

$\eta_{1} = 0.3$

$\eta_{2} = 0.6$

$T_{2} = 300\ \text{K}$

Solution:

Efficiency of Carnot engine:

$$\eta = \frac{T_{1} - T_{2}}{T_{1}} \quad (1),$$

where $T_{1}$ – the absolute temperature of the heater,

$T_{2}$ – the absolute temperature of the fridge

Of (1) $\Rightarrow \eta T_{1} = T_{1} - T_{2}$ (2)

Of (2) $\Rightarrow T_{1}(1 - \eta) = T_{2}$ (3)

Of (3) $\Rightarrow T_{1} = \frac{T_{2}}{1 - \eta}$ (4)

Of (4) $\Rightarrow T_{1}^{\prime} = \frac{T_{2}}{1 - \eta_{1}}$ (5)

Of (5) $\Rightarrow T_{1}^{\prime} = 429\ \text{K}$ (6)

Of (4) $\Rightarrow T_{1}^{\prime \prime} = \frac{T_{2}}{1 - \eta_{2}}$ (7)

Of (7) $\Rightarrow T_{1}^{\prime \prime} = 750\ \text{K}$ (8)

$\Delta T = T_{1}^{\prime \prime} - T_{1}^{\prime}$ (9)

(6) and (8) in (9): $\Delta T = 321\ \text{K}$

Answer:

321 K (321 °C)

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