A steel ball weighing 10 g is removed from the oven and immersed in water with a temperature of 10 s.

A steel ball weighing 10 g is removed from the oven and immersed in water with a temperature of 10 s. The water temperature rose to 25C. What was the temperature of the ball in the oven if the mass of water was 50g? Specific heat…

To find the temperature of the taken ball in the oven, we use the equality: Cc * mw * (tn1 – t) = Cw * mw * (t – tn2), whence we express: tn1 = Cw * mw * (t – tn2) / (Cc * mw ) + t.

Constants and variables: Cw – specific heat capacity of water (Cv = 4200 J / (kg * K)); mw is the mass of water (mw = 50 g = 0.05 kg); t – equilibrium temperature (t = 25 ºС); tн2 – initial water temperature (tн2 = 10 ºС); Сс – specific heat capacity of steel (Сс = 500 J / (kg * K)); msh is the mass of the steel ball (mw = 10 g = 0.01 kg).

Calculation: tn1 = Cw * mw * (t – tn2) / (Cc * mw) + t = 4200 * 0.05 * (25 – 10) / (500 * 0.01) + 25 = 655 ºС.