Question

If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter increases from 24.58°C to 35.82°C. Calculate the heat capacity of the calorimeter (in J/°C). The specific heat of water is 4.184 J/g•°C Group of answer choices

Answer 1

The value of the heat capacity of the Calorimeter
`C_c` = 54.4
`(J)/(c)`

Step-by-step explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water
`m_w` = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q =
`m_w C_w` ΔT +
`C_c` ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +
`C_c` × 11.24

611.37 =
`C_c` × 11.24

`C_c` = 54.4
`(J)/(c)`

This is the value of the heat capacity of the Calorimeter.