(a) Define specific heat capacity.
(b)(i) With the aid of a labelled diagram, describe an experiment to determine the specific heat capacity of copper using a copper ball.
(ii) State two precautions necessary to obtain accurate results.
(c) A piece of copper block of mass 24 g at 230°C is placed in a copper calorimeter of mass 60 g containing 54 g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water = 4200 J kg(^{-1}) K(^{-1})] [specific heat capacity of copper = 400 J kg(^{-1}) K(^{-1})]
Explanation
(a) Specific heat capacity is the quantity of heat energy required to change the temperature of unit mass (1kg) of a substance by 1k.
(b) Diagram
Method Weigh empty calorimeter plus stirrer and record the mass M(_1), weigh empty calorimeter plus stirrer plus water and record the mass m. Weigh copper ball and record the mass Mc. Take the initial temperature of the water T(_1). Heat copper ball in water to temperature T and record T. Quickly transfer the copper ball into the calorimeter and stir. Take the final temperature of the mixture T(_2). Heat lost by copper ball = Heat gained by calorimeter; water and stirrer.
McCc (T – T(_2)) = (M(_2) – M(_1))C(_w)
(T(_2) – T(_1)) + M(_1) Cal(T(_2) – T(_1))
Were Cc, Cw and Cz = specific heat capacity of copper, water and calorimeter respectively.
Cc = (frac{ (M – M_1C + M Cal (T – T)}{Mc(T – T_2)})
Precautions:
(i) Calorimeter should be lagged.
(ii) Quickly transfer the hot ball to the calorimeter.
(iii) Shake off water from hot calorimeter metal ball before transfer.
(iv) Mix thoroughly the mixture with stirrer.
(c) Heat lost = 0.024 x 400 (230 – T).
Heat gained by water = 0.054 x 4200 (T – 31)
Heat gained by calorimeter = 0.060 x 400 (T – 31)
Heat lost = Heat gained = 0.024 x 400 (230 – T) = 0.054 x 4200
(T – 31) + 0.060 x 400 (T – 31) T = 38.34°C