Question

What will be the value of temperature of hot reservoir in a carnot cycle, if the temperature of cold reservoir is 25°C heat absorbed from hot reservoir is 12J and heat rejected to cold reservoir is 4J?

• A. 894K
• B. 333.33K
• C. 684K
• D. 700K

Answer 1

Answer: (A)

894K

Answer 2

The problem asks us to find the temperature of the hot reservoir in a Carnot cycle, given the temperature of the cold reservoir and the heat absorbed and rejected.

1. Convert the temperature of the cold reservoir to Kelvin: The temperature of the cold reservoir, $T_c = 25^\circ C$. To convert Celsius to Kelvin, we add 273. $T_c = 25 + 273 = 298 K$

2. Apply the Carnot cycle relation between heat and temperature: For a Carnot cycle, the ratio of heat rejected to the cold reservoir ($Q_c$) to the heat absorbed from the hot reservoir ($Q_h$) is equal to the ratio of the absolute temperatures of the cold reservoir ($T_c$) and the hot reservoir ($T_h$). $\frac{Q_c}{Q_h} = \frac{T_c}{T_h}$

3. Substitute the given values and solve for $T_h$: Given: $Q_h = 12 J$ (heat absorbed from hot reservoir) $Q_c = 4 J$ (heat rejected to cold reservoir) $T_c = 298 K$

Substitute these values into the equation: $\frac{4 J}{12 J} = \frac{298 K}{T_h}$ $\frac{1}{3} = \frac{298 K}{T_h}$

Now, solve for $T_h$: $T_h = 3 \times 298 K$ $T_h = 894 K$

Thus, the temperature of the hot reservoir is 894 K.