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29 December, 23:29

A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process. What was the initial temperature of the ball? (ciron = 0.444 J/g°C)

A)

23°C

B)

78°C

C)

100°C

D)

155°C

+2
Answers (1)
  1. 29 December, 23:58
    0
    Heat lost or gained, H = mc (θ₂ - θ₁)

    Where m = mass, c = Specific heat capacity, θ₂ = final temperature, θ₁ = initial temperature

    m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).

    H = 6.9 kj = 6.9 * 1000J = 6900 J

    6900 = 200*0.444 * (θ₂ - 22)

    6900 / (200*0.444) = θ₂ - 22

    77.70 = θ₂ - 22

    θ₂ - 22 = 77.7

    θ₂ = 77.7 + 22 = 99.7

    So initial temperature before cooling ≈ 100°C. Option C.
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