Answer:
79%.
Step-by-step explanation:
Step 1:
Data obtained from the question. This include:
Input temperature = 100 °C.
Output temperature = 22 °C.
Step 2:
Conversion of celsius temperature to Kelvin temperature.
Temperature (Kelvin) = temperature (celsius) + 273
T (K) = T (°C) + 273
Input temperature = 100 °C.
Input temperature = 100 °C + 273 = 373 K.
Output temperature = 22 °C.
Output temperature = 22 °C + 273 = 295 K.
Step 3:
Determination of the efficiency of the locomotive.
Input temperature = 373 K.
Outpu temperature = 295 K.
Efficiency =...?
Efficiency = output /input x 100
Efficiency = 295/373 x 100
Efficiency = 79.1 ≈ 79%
Therefore, the efficiency of the locomotive is approximately 79%.