Question

in an experiment on the specific heat of a metal is 0.20 kg block of the metal at 150 degree celsius is dropped in a copper calorimetercontaining 150 cm cube of water at 27 degree Celsius the final temperature is 40 degree Celsius compare the specific heat of the metal if it losses to the surroundings are not ugly is in your answer generator or smaller than the actual value for specific heat of the metal ​

Answer 1

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

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

Hope it helps uhh ♥