Advertisement Remove all ads

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 is Your Answer Greater Or Smaller than the Actual Value for Specific Heat of the Metal - Physics

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 the specific heat of the metal?

Advertisement Remove all ads

Solution 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:

ΔT1 – T= 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 + mCw 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.

 

Solution 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

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 11 Thermal Properties of Matter
Q 14 | Page 295
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×