A 3kg piece of steel is heated to 100°C and placed into a calorimeter that contains 0.5kg of water at 20°C. The final temperature of the calorimeter was determined to be 52°C. What is the specific heat capacity of steel?
To solve this problem, we can use the principle of conservation of energy. The heat gained by the water and the calorimeter is equal to the heat lost by the steel.
The heat gained by the water and calorimeter can be calculated using the formula:
Q = mcΔT
Where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The heat lost by the steel can also be calculated using the same formula.
The heat gained by the water and calorimeter is:
Qwater = (0.5kg)(4186 J/kg°C)(52°C - 20°C) = 62790 J
The heat lost by the steel is:
Qsteel = (3kg)(c)(52°C - 100°C)
Since the heat gained is equal to the heat lost, we can set up the equation:
Qwater = Qsteel
62790 J = (3kg)(c)(52°C - 100°C)
Simplifying the equation:
62790 J = -48kg(c)
Dividing both sides by -48kg:
-1308 J/kg = c
Therefore, the specific heat capacity of steel is -1308 J/kg°C.