Question

A calorimeter of water equivalent 100 grams
contains 200 grams of water 10°C. A solid of
mass 500 grams at 45° C is added to the
Calorimeter. If equilibrium temperature is 25°C
then, the specific heat of the solid is
(A) 0.45 cal g-10c-1
(B) 0.1cal g-10c-1
(C) 4.5cal g-100-
(D) 0.01 cal g=10c-1​

Answer 1

Answer is........

A)0.45 cal g-10 c-1

Answer 2

Given info : A calorimeter of water equivalent 100 grams contains 200 grams of water 10°C. a solid of mass 500 grams at 45° C is added to the

Calorimeter.

To find : if equilibrium temperature is 25°C then find the specific heat capacity of the solid.

solution : total mass of water and water equivalent, m = 200g + 100g = 300g

specific heat of water , s = 1 cal/g/°C

change in temperature, ∆T = (25°C - 10°C) = 15°C

so, heat gained by water = ms∆T

= 300g × 1 cal/g/°C × 15°C

= 4500 cal

mass of solid, M = 500g

change in temperature, ∆T' = (45°C - 25°C) = 20°C

let specific heat capacity of the solid is s'

so, heat lost by solid = Ms'∆T'

= 500g × s' × 20°C .....(2)

from calorimetry,

heat gained = heat lost

4500 = 500 × s' × 20

⇒s = 4500/10000 = 0.45cal/g/°C

Therefore the specific heat of the solid is 0.45 cal/g/°C