Question

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 cm3of 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?
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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, Cw = 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 CwΔT’

= (M + m′) Cw ΔT’ … (ii)

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

mCΔT = (M + m’) Cw ΔT’

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

∴ C = (175 × 4.186 × 13) / (110 × 200) = 0.43 Jg-1K-1

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

Answer 2

ʜᴇʀᴇ ɪs ʏᴏᴜʀ ᴀɴsᴡᴇʀ

Here, mass of metal block ( m1) = 0.2 Kg

Temperature of metal block ( T1) = 150°C

Temperature of calorimeter ( T2) = 27°C

Final temperature of the mixture ( T) = 40° C

Volume of water in calorimeter = 150 cm³

Mass of water in calorimeter = volume × density

= 150 × 10^-6 m³ × 10³ × kg/m³

= 150 × 10^-3 kg

Water equivalent of calorimeter = 0.025 kg = 25 × 10^-3 kg

Let s is the specific heat of the metal block .

Use formula,

Q = mS∆T

so, Q1 = m1S ( T1 - T2)

= 0.2 × S × ( 150 - 40)

= 22S

Now,

Heat taken by calorimeter and water

Q2 = (150 × 10^-3 + 25 × 10^-3)×S2 ×(40-27)

= 175 × 10^-4 × 4.2 × 10³ × 13

= 175 × 4.2 × 13

A/c to Calorimeter ,

Q1 = Q2

22S = 175 × 4.2 × 13

S = 175 × 4.2 × 13/22

= 434 J/kg-K

NOTE:-  If heat loses to the surrounding are not negligible than the value of specific heat 'C' of metal will be less than actual value .

ᴇʜsᴀss   ^_^)!!