Question

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 94.8 g of water at 60.4°C with 94.8 g of water, already in the calorimeter, at 22.3°C, the final temperature of the water is 35.0°C. Calculate the heat capacity of the calorimeter in J/K. Use 4.184 J/g°C as the specific heat of water. Enter to 1 decimal place.

Answer 1

Answer:

$396.65$ JC⁻¹

Explanation:

$m_{a}$ = mass of water added to calorimeter = 94.8 g

$T_{ai}$ = initial temperature of the water added = 60.4 C

$c_{w}$ = specific heat of water = 4.184 Jg⁻¹C⁻¹

$m_{c}$ = mass of water available to calorimeter = 94.8 g

$T_{ci}$ = initial temperature of the water in calorimeter = 22.3 C

$T_{f}$ = final equilibrium temperature = 35 C

$Q_{C}$ = Heat gained by calorimeter

Using conservation of heat

Heat gained by calorimeter = Heat lost by water added - heat gained by water in calorimeter

$Q_{C} = m_{a} c_{w} (T_{ai} - T_{f}) - m_{c} c_{w} (T_{f} - T_{ci})$

$Q_{C} = (94.8) (4.184) (60.4 - 35) - (94.8) (4.184) (35 - 22.3)$

$Q_{C} = 5037.4$ J

$T$ = Change in temperature of calorimeter

Change in temperature of calorimeter is given as

$T = T_{f} - T_{ci}$

$T = 35 - 22.3$

$T = 12.7$ C

Heat capacity of calorimeter is given as

$c_{cm} = \frac{Q_{C}}{T}$

$c_{cm} = \frac{5037.4}{12.7}$

$c_{cm} = 396.65$ JC⁻¹