Answer:
79%
Explanation:
Data obtained from the question include the following:
Input temperature = 100°C
Output temperature = 22°C
Efficiency =.?
Next, we shall convert celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T (°C) + 273
Input temperature = 100°C
Input temperature in Kelvin = 100°C + 273 = 373K
Output temperature = 22°C
Output temperature in Kelvin = 22°C + 273 = 295K
Finally, we shall determine the efficiency of the locomotive as follow:
Efficiency = output /input x 100
Output = 295K
Input = 373K
Efficiency =?
Efficiency = output /input x 100
Efficiency = 295/373 x 100
Efficiency = 79%
Therefore, the efficiency of the locomotive is 79%.
