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 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 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 .

Answer 2

Answer:

0.43kJ/kgK smaller than value of specific heat of the metal

Explanation:

refer the attachment.