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Which is a correct statement about an ideal heat engine? Tick (✓) one box - AQA - A-Level Physics - Question 5 - 2021 - Paper 6

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Which is a correct statement about an ideal heat engine? Tick (✓) one box. The efficiency is increased when the kelvin temperatures of the hot source and the cold ... show full transcript

Worked Solution & Example Answer:Which is a correct statement about an ideal heat engine? Tick (✓) one box - AQA - A-Level Physics - Question 5 - 2021 - Paper 6

Step 1

The efficiency is 50% when the kelvin temperature of the hot source is twice the kelvin temperature of the cold sink.

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Answer

This statement is correct according to the efficiency equation for an ideal heat engine given by:

extEfficiency=1TcTh ext{Efficiency} = 1 - \frac{T_c}{T_h}

If the hot source temperature ThT_h is twice the cold sink temperature TcT_c, then: Th=2TcT_h = 2T_c Substituting this into the efficiency equation gives: extEfficiency=1Tc2Tc=112=0.5extor50% ext{Efficiency} = 1 - \frac{T_c}{2T_c} = 1 - \frac{1}{2} = 0.5 ext{ or } 50\%

Step 2

Determine the coefficient of performance COP_ref of the refrigerator.

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Answer

The coefficient of performance for a refrigerator, COP_ref, is defined as:

COPref=QhWCOP_{ref} = \frac{Q_h}{W}

We know: W=QhQcW = Q_h - Q_c

For an ideal engine efficiency given as 0.330.33: η=WQh=0.33\eta = \frac{W}{Q_h} = 0.33

Therefore, substituting for work: W=0.33QhW = 0.33 Q_h

Now rearranging gives: Qc=QhW=Qh0.33Qh=0.67QhQ_c = Q_h - W = Q_h - 0.33 Q_h = 0.67 Q_h

Substituting this back into the COP formula yields: COPref=Qh0.33Qh=10.333.03COP_{ref} = \frac{Q_h}{0.33 Q_h} = \frac{1}{0.33} \approx 3.03

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