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 cm of
water at 27° C. The final temperature is 40°C.
Calculate the specific heat of metal.
(1) 0.99 J/K
(2) 1.26J/g/K
(3) 0.43 J/g/K (4) - 1.26 J/g/K
A substance of n
Answers
Answer:
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 .
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