Answer:571.09 kJ
Explanation:
Given
Temperature of cooling water from engine exit[tex]=240^{\circ} F\approx 115.55^{\circ}C[/tex]
After Passing through the radiator its temperature decreases to [tex]175^{\circ}F\approx 79.44^{\circ}F[/tex]
specific heat of water[tex]=4.184 J/g^{\circ}C[/tex]
Volume of water [tex]= 1 gallon\approx 3.78 L[/tex]
density of water [tex]\rho =1 gm/mL[/tex]
Thus mass of water[tex]=\rho \times V=3.78\times 1=3.78 kg [/tex]
Heat transferred to the surrounding is equal to heat absorbed by cooling water
[tex]Q=m\cdot c\cdot \Delta T[/tex]
[tex]Q=3.78\times 4.184\times 1000\times (115.55-79.44)[/tex]
[tex]Q=3.78\times 4.184\times 1000\times (36.11)[/tex]
[tex]Q=571.09 kJ[/tex]
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 is 572645.67 J
Heat: This can be defined as a form of energy that brings about the sensation of warmth. The S.I unit of heat is Joules (J)
The Question above can be solved using the formula below
Q = cm(t₁-t₂)............... Equation 1
Where Q = amount of heat, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: c = 4.184 J/g°C, m = 1 gallon = 3785 g, t₁ = 240 °F = 115.56 °C, t₂ = 79.4 °C
Substitute these values into equation 1
Q = 4.184(3785)(115.56-79.4)
Q = 572645.67 J
Hence the amount of heat transferred from the engine to the surrounding is 572645.67 J
Learn more about heat here: https://brainly.com/question/21927103
