Question

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)

Answer 1

It's 100°C or C) on USATestprep

Step-by-step explanation:

Answer 2

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