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

The value of  the heat capacity of the Calorimeter   = 54.4

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water = 75.40 gm

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

From the given condition

Q = ΔT + ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +   × 11.24

611.37 =   × 11.24

= 54.4

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

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