Question

A solid of mass 0.15 kg and at 100°C is placed in 0.25 kg of water, contained in a copper calorimeter of mass 0.12 kg, at 10°C. If the final temperature of the mixture is 20°C, calculate the sp. heat capacity of the solid.

[Given S.H.C. of water = 4200 Jkg⁻¹ K⁻¹; S.H.C. of copper = 400 Jkg⁻¹ K⁻¹]

Answer 1

Given: m_s = 0.15 kg, θ_s = 100°C

m_w = 0.25 kg, m_c = 0.12 kg, θ_w = θ_c = 10°C

c_w = 4200 Jkg⁻¹ K⁻¹, c_c = 400 Jkg⁻¹ K⁻¹, θ = 20°C

Fall in temperature of the solid:

θ_Fs = 100 − 20 = 80°C

Rise in temperature of water and calorimeter:

θ_R = 20 − 10 = 10°C

Heat lost by the solid:

Q_s = m_sc_sθ_Fs

= 0.15 × c_s × 80

= 12c_s J

Heat gained by water:

Q_w = m_wc_wθ_R

= 0.25 × 4200 × 10

= 10500 J

Heat gained by copper calorimeter:

Q_c = m_cc_cθ_R

= 0.12 × 400 × 10

= 480 J

Total heat gained:

Q_g = 10500 + 480 = 10980 J

By the principle of calorimetry,

Q_s = Q_g

12c_s = 10980

$$c_s=\frac{10980}{12}$$

= 915 Jkg⁻¹K⁻¹

Therefore, the specific heat capacity of the solid is 915 Jkg⁻¹K⁻¹.