300 cal. of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be _______°C.
To solve this problem, we need to determine the temperature of the sink (\(T_2\)) using the known values and the principles of heat engines. From thermodynamics, the efficiency (\(\eta\)) of a heat engine is given by: \(\eta = \frac{Q_1-Q_2}{Q_1}\) where:
\(Q_1 =\) heat absorbed (in calories) = 300 cal
\(Q_2 =\) heat rejected (in calories) = 225 cal
The efficiency can also be expressed in terms of temperatures of the source (\(T_1\)) and sink (\(T_2\)) as: \(\eta = \frac{T_1-T_2}{T_1}\) (using absolute temperature in Kelvin). First, calculate the efficiency using the heat values: \(\eta = \frac{300-225}{300} = \frac{75}{300} = 0.25\) Convert the source temperature from Celsius to Kelvin: \(T_1 = 227 + 273 = 500 \text{ K}\) Using the formula for efficiency: \(\frac{T_1-T_2}{T_1} = 0.25\) Solve for \(T_2\): \(T_1-T_2 = 0.25 \times T_1\) which means: \(T_2 = T_1 - 0.25 \times T_1 = 500 - 0.25 \times 500 = 500 - 125 = 375 \text{ K}\) Convert back to Celsius for \(T_2\): \(T_2 = 375 - 273 = 102^\circ\text{C}\) The calculated sink temperature \(T_2\) is 102°C, which falls within the expected range of 102 to 102°C.
