A 10.5-g sample of ethylene glycol, a car radiator coolant, loses 546 J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5°C? (c of ethylene glycol = 2.42 J/g·K)

the initial temperature is T initial = 53.5 °C

Explanation:

the heat lost Q by ethylene glycol is

Q = m * c * (T final - T initial)

where

m = mass of ethylene glycol = 10.5 g

c = specific heat capacity of ethylene glycol = 2.42 J/g*°C

T final = final temperature = 32.5°C

T initial = initial temperature

Q = heat lost = (-546 J) (negative sign means lose of energy)

solving for T initial

Q = m * c * (T final - T initial)

T initial = T final - Q / (m * c)

replacing values

T initial = T final - Q / (m * c) = 32.5°C - (-546 J) / (10.5 g * 2.42 J/g*°C) = 53.5 °C

T initial = 53.5 °C

then the initial temperature is T initial = 53.5 °C