A reversible heat engine operates with an efficiency of 50%. If during each cycle it rejects 150 cal to a reservoir of heat at 30°C, then what is the temperature of the other reservoir and how much work does it carry out per cycle?
To find the temperature of the other reservoir, we can use the efficiency formula for a heat engine:
Efficiency = 1 - (Tc/Th)
Where Tc is the temperature of the cold reservoir (30°C) and Th is the temperature of the hot reservoir that we need to find.
Given that the efficiency is 50%, we have:
0.5 = 1 - (30/Th)
Rearranging the equation, we get:
30/Th = 1 - 0.5
30/Th = 0.5
Cross-multiplying, we get:
Th = 30 / 0.5
Th = 60°C
Therefore, the temperature of the other reservoir is 60°C.
To find the work carried out per cycle by the heat engine, we can use the work formula for a heat engine:
Work = Qh - Qc
Where Qh is the amount of heat absorbed from the hot reservoir and Qc is the amount of heat rejected to the cold reservoir.
Given that Qc is 150 cal and Qh is the heat absorbed from the hot reservoir, we only need to find Qh to calculate the work.
Given that the efficiency is 50%, we have:
Efficiency = Work / Qh
0.5 = Work / Qh
Cross-multiplying, we get:
Work = 0.5 * Qh
Since the heat engine is reversible, Qh is equal to the amount of heat rejected to the cold reservoir. Therefore, Qh is also 150 cal.
Plugging in the values, we have:
Work = 0.5 * 150 cal
Work = 75 cal
Therefore, the heat engine carries out 75 calories of work per cycle.
To find the temperature of the other reservoir, we can use the formula for the efficiency of a heat engine:
Efficiency = 1 - (Tc/Th),
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
Given that the efficiency is 50% (or 0.5), we can substitute this value into the formula:
0.5 = 1 - (Tc/Th).
Rearranging the equation, we get:
Tc/Th = 0.5.
We are also given that 150 calories of heat is rejected to the cold reservoir during each cycle. Using the formula for heat, Q = mcΔT, where Q is the heat absorbed or rejected, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature, we have:
150 cal = mcΔT.
Since we are given only the heat and the change in temperature, we can rewrite the equation as:
150 cal = mc(Tc - 30°C).
Now, let's find the work done per cycle. The work done by a heat engine is given by the formula:
Work = Qh - Qc,
where Qh is the heat absorbed from the hot reservoir and Qc is the heat rejected to the cold reservoir.
In this case, Qc is given as 150 cal, and since the heat engine is reversible, we know that Qh = Qc.
Thus, the work done per cycle is:
Work = Qc - Qc = 0.
Therefore, the heat engine does not carry out any work per cycle.
Now, let's solve for the temperature of the other reservoir and the temperature of the cold reservoir:
Tc/Th = 0.5.
We can assign a value to Th, let's say Th = 100°C (since the temperature of the cold reservoir is given as 30°C).
Plugging in the values, we have:
Tc/100 = 0.5.
To solve for Tc, we can cross-multiply:
Tc = 0.5 * 100.
Tc = 50°C.
So, the temperature of the other reservoir is 50°C. And the heat engine does not carry out any work per cycle.
.5=(tin-tout)/tin
where tout is (30+273) , solve for tin (temp of original reservoir)
work per cycle=300cal (to get 50 percent efficiency)