300 grams of ethanol is heated with 14640 Joules of energy to reach a final temperature of 30 °C. What was the initial temperature of the ethanol? The specific heat of ethanol is 2.44 J/g·°C.

10 °C

Explanation:

Applying

q = cm(t₂-t₁) Equation 2

Where q = heat energy, c = specific heat of ethanol, m = mass of ethanol, t₁ = initial temperature, t₂ = Final temperature.

Given: c = 2.44 J/g.°C,  m = 300 g, q = 14640 J, t₂ = 30°C

Substitute into equation 2 and solve for t₁

14640 = 2.44×300(30-t₁)

14640 = 732(30-t₁)

732(30-t₁) = 14640

(30-t₁)  = 14640/732

(30-t₁)  = 20

t₁ = 30-20

t₁ = 10 °C

30°C

Explanation:

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