Answer 1

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T₁ = 150°C

Final temperature of the metal, T₂ = 40°C

Calorimeter has water equivalent of mass, m^’ = 0.025 kg = 25 g

Volume of water, V = 150 cm³

Mass (M) of water at temperature T = 27°C:

150 × 1 = 150 g

Fall in the temperature of the metal:

ΔT = T₁ – T₂ = 150 – 40 = 110°C

Specific heat of water, C_w = 4.186 J/g/°K

Specific heat of the metal = C

Heat lost by the metal, θ = mCΔT … (i)

Rise in the temperature of the water and calorimeter system:

ΔT′^’ = 40 – 27 = 13°C

Heat gained by the water and calorimeter system:

Δθ′′ = m₁ C_wΔT^’

= (M + m′) C_w ΔT^’ … (ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔT = (M + m^’) C_w ΔT^’

200 × C × 110 = (150 + 25) × 4.186 × 13

`:. C = (175xx4.186xx13)/(110xx200) = 0.43 J g^(-1) K^(-1)`

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

 

 

Answer 2

 Mass of metal block, m = 0.20 kg = 200 g

Fall in the temperature of metal block,

ΔT = (150 – 40) °C = 110 °C

If C be the specific heat of metal, then heat lost by the metal block = 200 x C x 110 cal Volume of water = 150 cm³

mass of water = 150 g

Increase in temperature of water = (40 – 27) °C = 13°C

Heat gained by water = 150 x 13 cal Water equivalent of calorimeter, w = 0.025 kg = 25g

Heat gained by calorimeter, `"w x increase in temperature of calorimeter"`

= 25 x 13 cal

Heat lost by metal block =  Heat gained by water  +  Heat gained by calorimeter

200 x C x 110 = (150 + 25 ) 13

`C = (175xx13)/(200xx100) =  0.1 Cal g^(-1) ""^@C^(-1) = 0.43 J g^(-1) K^(-1)`

if heat is lost to the surroundings, C will be smaller then the actual value