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

Answer:

The value of  the heat capacity of the Calorimeter  $C_c$ = 54.4 $\frac{J}{c}$

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 $\frac{J}{c}$

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