A piece of copper weighing 400g is heated to 100C and then quickly transferred to a copper calorimeter of mass 10g containing 100g of liquid of unknown specific heat capacity at 30C. If the final temperature of mixture is 50C, calculate the specific heat capacity of the liquid.
First, we need to calculate the heat gained by the liquid and calorimeter:
Q = mcΔT
For the liquid and calorimeter:
Q = (100g + 10g) * c * (50 - 30)
Q = 110g * c * 20
Q = 2200c
Next, we need to calculate the heat lost by the copper:
Q = mcΔT
Q = 400g * 0.385 J/g°C * (Tf - 100)
Since the heat lost is equal to the heat gained:
400 * 0.385 * (Tf - 100) = 2200c
Solving for Tf:
Tf - 100 = 2200c / 154
Tf = 100 + 14.29c
Substitute Tf back into the equation:
400 * 0.385 * (100 + 14.29c - 100) = 2200c
153.9c = 2200c
153.9 = 2200
c = 14.29 J/g°C
Therefore, the specific heat capacity of the liquid is 14.29 J/g°C.