The temperature of the cooling water as it leaves the hot engine of an automobile is 240°F. After it passes through the radiator it has a temperature of 175°F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g°C.

Under ideal circumstances, the net heat change is zero.

\({q_{subs\tan ceM}} + {q_{subs\tan ceN}}\) = 0

M is equal to the heat lost by the substance W.

\({q_{subs\tan ceM}} = - {q_{subs\tan ceN}}\)

The negative sign merely shows that the direction of the heat flow is opposite to each other.

Given information:

The density of the water = 1g/mL

The volume of the water = 1gallon = 3785.412mL

Therefore, the mass of the water = density \( \times \) volume = 3785.412\( \times \)1 = 3785.412g

The change in temperatures is given in Fahrenheit, so we need to convert it to Celsius.

Change in temperature = \({T_{final}} - {T_{initial}} = \) 79.44 – 115.55 = -36.11\(^0C\)

\({T_{initial}} = \)\(^0C = \frac{5}{9} \times {(^0}F - 32)\)= \(^0C = \frac{5}{9} \times {(^0}F - 32) = \frac{5}{9} \times (240 - 32) = {115.55^0}C\)

\({T_{final}} = \)\(^0C = \frac{5}{9} \times {(^0}F - 32)\)= \(^0C = \frac{5}{9} \times {(^0}F - 32) = \frac{5}{9} \times (175 - 32) = {79.44^0}C\)

The specific heat of the water = 4.186\(\frac{J}{{{g^0}C}}\)

Therefore, Q = \(3785.412 \times 4.186 \times ( - 36.11) = - 627491J = - 627.5kJ\)

The negative sign implies that the heat energy is released in the process.

We know that \({q_{subs\tan ceM}} = - {q_{subs\tan ceN}}\).

The heat transferred by the engine to the surrounding = 627.5 kJ.