Question

Calculate the calorimeter constant in trial one.

Data: Number Unit Trial 1 50 mL Mass of cold water 23.1 C Temperature of cold water 50.0 mL Mass of hot water 51.8 C Temperature of hot water 37.7 C Temperature of mixed water Number Unit 0.105 g Mg (s)HCI (aq) Mass of Mg 100.0 mL Volume of HC 22.4 C Temperature of HC

Answer 1

Here the energy is gain by cold water and it is lost by hot water. Therefore

Heat lost by hot water = Heat gain by cold water

Q_hot = Q_cold

Here mass of water is same for both water i.e. 50 g because density of water is 1.00 g/ml and mass is calculated by

d=m/v

1.00 g/ml = m/50 ml

m= 1.00 g/ml * 50 ml = 50 g.

50.0 g * 4.184 J/g^oC * (37.7-51.8) = 50.0 g * 4.184 J/g^oC * (37.7-23.1)

50.0 g * 4.184 J/g^oC * (-14.1) ^oC = 50.0g * 4.184 J/g^oC * 14.6 ^oC

2949.72 J = 3054.32 J

Therefore heat got by calorimeter is

3054.32 - 2949.72 = 104.6 J

The calorimeter constant is calculated by

Q(calorimeter) / T(cold water)

= 104.6 / 14.6 = 7.1643 J/^oC

= 7.16 J/^oC

Therefore calorimeter constant is 7.16 J/^oC.

If you find any mistake, please mention in comment box.

Thanks.