A heat engine operating between a cold reservoir temperature of 300K and a hot reservoir takes 200J of heat from the hot reservoir and delivers 120J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir is :
as u know .... efficiency=W/Qhot x 100 ............ W=Qhot-Qcold........................ so from here u can get the value of work and put it into the above efficiency equation ......... after getting the answer of efficiency then apply formula ................... efficiency=1-Tcold/Thot x 100
To determine the minimum temperature of the hot reservoir, we can use the Carnot efficiency formula:
Efficiency = 1 - (Tc/Th)
Where:
- Efficiency is the ratio of work output to heat input.
- Tc is the temperature of the cold reservoir.
- Th is the temperature of the hot reservoir.
Given:
- Tc = 300K (cold reservoir temperature)
- Qh = 200J (heat taken from hot reservoir)
- Qc = 120J (heat delivered to cold reservoir)
First, let's calculate the efficiency:
Efficiency = 1 - (Qc/Qh)
= 1 - (120/200)
= 1 - 0.6
= 0.4
Now, let's rearrange the Carnot efficiency formula to find the temperature of the hot reservoir:
Efficiency = 1 - (Tc/Th)
0.4 = 1 - (300/Th)
Solving for Th:
0.4 = 1 - (300/Th)
0.4 - 1 = -300/Th
0.6 = 300/Th
Th = 300/0.6
Th = 500 K
Therefore, the minimum temperature of the hot reservoir is 500K.