Question

What is the heat capacity of an object at 25.5∘C that absorbs 45 kJ of heat and is heated to 28.2∘C?

Answer 1

Answer:

16.6 kJ/°C

Explanation:

given,

Amount of heat absorbed = 45 kJ

initial temperature, T₁ = 25.5°C

final temperature, T₂ = 28.2°C

change in temperature = T₂ - T₁

                                       = 28.2 - 25.5  = 2.7° C

$Heat\ capacity = \dfrac{Heat\ absorbed}{\Delta T}$

$Heat\ capacity = \dfrac{45\ kJ}{2.7}$

$Heat\ capacity = 16.6\ kJ/^0C$

Heat capacity of the object is equal to 16.6 kJ/°C

Answer 2

Answer:

16666.67 J/°C

Explanation:

t1 = 25.5 °C

T2 = 28.2 °C

heat, H = 45 kJ = 45000 J

the amount of heat required to raise the temperature of substance by 1 °C is called heat capacity.

Heat capacity = heat / rise in temperature

heat capacity = 45000 / ( 28.2 - 25.5)

heat capacity = 16666.67 J/°C

Thus, the heat capacity of the substance is 16666.67 J/°C.