Question

A hot solid of mass 60g at 100°C is placed in 150 g of water at 20°C. The final steady temperature recorded is 25°C. Calculate the specific heat capacity of the solid.
[Specific heat capacity of water = 4200 J kg-1 °C-1] ​

Answer 1

Explanation:

Given : Mass of hot solid, m = 60 gm, at 1000C

Mass of water, m2 = 150 gm, at 200 C

Final temperature = 250C.

By principle of calorimetry,

Heat given = Heat taken

⇒ m1c1ΔQ1 = m2c2ΔQ2

⇒ 60 x c x (100 – 25) = 150 x 4.2 x (25 – 20)

⇒ ∴ Required specific heat, c1 = 0.7J/gm 0 C

Hope it's helpful to you

Answer 2

Answer:

Heat gained = Heat lost

150×4.4×(25−50)=60×c×(100−25)

150×4.2×5=60×c×75

Specific heat capacity (c)=0.7Jg