In an experiment on the specific heat of a metal, a 0.20 kg block of

1.	In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Answer» Mass of the block, m₁ = 0.20 kg = 0.02 x 1000 = 200 gm Temperature of the block, θ₁ = 150°C Mass of water in the calorimeter, m₂ = Volume x density = 150 x 1g = 150g Water equivalent, W = 0.025 Kg = 0.025 x 1000 = 25 kg Temperature of calorimeter, θ₂ = 27°C Specific heat capacity of water, C₂ = 1 cal g^-1 = 4.184 Jg^-1 Let C₁ be the specific heat capacity of the block. Let θ be the final temperature. Then θ = 40°C. Using the principle of calorimetry, we get m₁C₁ (θ₁ - θ) = (m₁ + ω) C₂(θ - θ₂) ⇒ C₁ = {(m₂ + ω) C₂(θ - θ₂)}/{m₁(θ₁ - θ)} = {(150 + 25) x 4.184 x (40 - 27)}/{200 (150 - 40)} = 0.43 Jg^-1K^-1 If the heat losses to the surrounding are not negligible, the answer will be smaller compared to 0.43 Jg^-1K^-1.

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Answer»

Mass of the block, m₁ = 0.20 kg

= 0.02 x 1000 = 200 gm

Temperature of the block, θ₁ = 150°C

Mass of water in the calorimeter,

m₂ = Volume x density = 150 x 1g = 150g

Water equivalent,

W = 0.025 Kg = 0.025 x 1000 = 25 kg

Temperature of calorimeter, θ₂ = 27°C

Specific heat capacity of water,

C₂ = 1 cal g^-1 = 4.184 Jg^-1

Let C₁ be the specific heat capacity of the block. Let θ be the final temperature. Then θ = 40°C. Using the principle of calorimetry, we get

m₁C₁ (θ₁ - θ) = (m₁ + ω) C₂(θ - θ₂)

⇒ C₁ = {(m₂ + ω) C₂(θ - θ₂)}/{m₁(θ₁ - θ)}

= {(150 + 25) x 4.184 x (40 - 27)}/{200 (150 - 40)}

= 0.43 Jg^-1K^-1

If the heat losses to the surrounding are not negligible, the answer will be smaller compared to 0.43 Jg^-1K^-1.