Ahot 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)

99.7 °C

Step-by-step explanation:

The units of ciron tell us that in order to have °C in the numerator, we need to divide the heat loss by the product of the mass and ciron:

∆T = (6900 J)/(0.444 j/g°C × 200 g) = 69/0.888 °C ≈ 77.7 °C

This is the change in temperature as the ball cooled, so its initial temperature was ...

22 °C +77.7 °C = 99.7 °C

my acc plz

step-by-step explanation:

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation: