Question

50 ml of hot water was mixed with 50ml of cold water in a calorimeter. The initial temperature of the cold water is 22°C, the initial temperautre of hot water is 79°C, the average temperature is 50.5°C, the temperature at time 0 is 46.333°C.

What is the heat lost by hot water, the heat gained by cold water, the heat gained by calorimeter in J, temperature change of the calorimeter, the heat capacity (C) of the calorimeter in J/°C, and the average heat capacity of calorimeter in J/°C? Please show work

Answer 1

Density of water = 1 g/mL

Mass of hot water = density x volume = 1 x 50 = 50 g

Mass of cold water = density x volume = 1 x 50 = 50 g

Initial temperature of cold water = 22 °C

Final temperature of cold water = 46.33 °C

Temperature change in cold water = 46.33 - 22= 24.33 °C

Initial temperature of hot water = 79 °C

Final temperature of hot water = 46.33 °C

Temperature change in hot water = 79 - 46.33 = 32.67 °C

Specific heat capacity of water Cpw = 4.184 J/gC

Heat capacity of the calorimeter = C J/°C

Part a

Heat lost by hot water = mass of hot water x Cpw x temperature change in hot water

= 50 g x 4.184 J/gC x 32.67 C

= 6834.564 J

Part b

Heat gained by cold water = mass of cold water x Cpw x temperature change in cold water

= 50 g x 4.184 J/gC x 24.33 C

= 5089.836 J

Part C

Heat gained by calorimeter = heat lost by hot water - heat gained by cold water

= 6834.564 - 5089.836 = 1744.728 J

Part d

temperature change of the calorimeter = Temperature change in cold water

= 46.33 - 22 = 24.33 °C

Part e

Heat gained by calorimeter = C x temperature change of the calorimeter

1744.728 J = C x 24.33 °C

C = 71.71 J/°C