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