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 incr

Question

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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 incr

eases 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

Chemistry

1 answer:

VMariaS [2.9K] · Answer 1 · 2026-03-10T15:37:58+03:00

Answer:

The heat capacity of the calorimeter is $C_c$ = 54.4 $\frac{J}{c}$

Explanation:

Given the data

Heat supplied Q = 4.168 KJ = 4168 J

Mass of water $m_w$ = 75.40 gm

Change in temperature = ΔT = 35.82 - 24.58 = 11.24 °C

From the conditions provided

Q = $m_w C_w$ ΔT + $C_c$ ΔT

Plugging all values into the above equation yields

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 represents the heat capacity of the calorimeter.

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 incr

Answer;

Explanation;