A 10-kg iron ball is dropped onto a pavement from a height of 100 m. If half of the heat generated goes into 3 warming the ball, find the temperature increase of the ball. (In SI units, the specific heat capacity of iron is 450 J/kg°C.) Why is the answer the same for a ball of any mass?
Solution 7P In this case the heat generated by the ball comes from the potential energy of the ball. Step 1 The potential energy of the ball is given by E P mgh Where m is the mass of the ball, g is the acceleration due to gravity and h is the height of the ball. From the given question we know that m = 10 kg g = 9.8 m/s2 h = 100 m 2 So the potential energy ispE = (10 kg)(9.8 m/s )(100 m) = 9800 J Half of the energy will be used to increase the temperature. Hence the energy that will be used to increase the temperature is 9800 J Q = 2 = 4900 J Step 2 In the increase in temperature is T then we know that Q = mcT Q T = mc Here we have m = 10 kg and c = 450 J/kg°C Using putting this values in the above equation we have T = (10 kg)(450 J/kg°C)09°C Hence the temperature of the iron ball will increase by 1.09°C.