Thermal Properties of Matter Numerical on Specific Heat, Latent Heat and Calorimetry Class-11 Nootan Physics Solutions Ch-16. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Thermal Properties of Matter Numerical on Specific Heat, Latent Heat and Calorimetry Class-11 Nootan Solutions
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-17 | Thermal Properties of Matter |
| Topics | Numerical on Specific Heat, Latent Heat and Calorimetry |
| Academic Session | 2024-2025 |
Solved Numericals Based on Specific Heat, Latent Heat and Calorimetry
(Thermal Properties of Matter Numerical-2nd Part ISC Class- Nootan Physics Ch-16)
Que-12: An aluminium sphere of mass 0.047 kg is heated to 100°C. It is dropped in a copper calorimeter of mass 0.14 kg, containing 0.25 kg of water at 20°C. The temperature of water rises to a steady state at 23°C. Calculate the specific heat of aluminium. Specific heat of water 4.18 x 10³ J kg^-1 °C^-1, specific heat of copper = 0.386 × 10³ J kg^-1 °C^-1
Ans : According to the principle of Calorimetry
=> Heat given = Heat Taken
=> 0.047 x C(AC) x (100 – 23)
=> 0.25 x 4.18 x 10³ (23 – 20)
=> C(AC) = 0.25 x 4.18 x 10³ x 3 / 0.047 x 77
=> 911 J Kg^-1 °C^-1
Que-13: A calorimeter contains 75 g of water at 15°C. When 50 g water of 100°C is poured in the calorimeter, the temperature of the mixture becomes 25°C. Calculate water equivalent of the calorimeter.
Ans: Let water equivalent of calorimeter is m gram then,
=> Heat given = Heat taken
=> 50 x c x (100 – 25) = (75 + m)c (25 – 15)
=> 75 + m = 50 x 75
=> 75 + m = 375 m
=> m = 375 – 75 = 300 g
Que-14: When 0.15 kg of ice at 0°C is mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7 °C. Calculate latent heat of melting of ice. Specific heat of water is 4.186 × 10³ J kg^-1 K^-1
Ans: By principle of Calorimetry
=> Heat given = Heat Taken
=> 0.30 x 4.186 x 10³ x (50 – 6.7)
=> 0.15 x L + 0.15 x 4.186 x 10³ x 6.7
=> 0.15 L = 0.3 x 4.18 x 10³ – 0.15 x 4.186 x 10³ x 6.7
=> 4.186 x 10³ (0.3 x 43.3 – 0.15 x 6.7)
=> L = 4.186 x 10³ x 12 / 0.15
=> 3.34 x 10^5 J Kg^-1
Que-15: Calculate the heat required to convert 3 kg of ice at -12°C kept in a calorimeter to steam at 100° at atmospheric pressure. Given : specific heat of ice = 2.100 x 10^5 J kg^-1 K^-1, specific heat of water = 4.186 x 10³ J kg^-1 K^-1, latent heat of fusion of ice = 3.35 x 10^5 J kg^-1 and latent heat of steam = 2.256 x 10^6 J kg^-1.
Ans-15 Total heat required
=> 3 x (0 – (-12) x 2.1 x 10^5 + 3 x 3.5 x 10^5 + 3 x (100 – 0) x 4.2 x 10^5 + 3 x 22.56 x 10^5
=> 9.1 x 10^6 J
Que-16: The temperature of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature, when A and B are mixed, is 16°C, and when B and C are mixed, is 23°C. What will be the temperature when A and C are mixed ?
Ans: Let m be the mass of each liquid. Let the specific heats of A, B and C be s1, s2 and s3 respectively.
(i) When A and B are mixed:
Heat gained by A = Heat lost be B
=> ms1 (16 − 12) = ms2 (19 – 16)
or s2 = 4/3 s1 ………….(i)
(ii) When B and C are mixed:
Heat gained by B = Heat lost be C
= ms2 (23 -19) = ms3 (28 – 23)
= 4 S2 = 5 S3
∴ s3 = 4/5 s2 = 4/5 x 4/3s1 = 16/15 s1 ……….(ii)
(iii) When A and C are mixed:
Let be the resultant temperature.
Heat gained by A = Heat lost by C
ms1 (θ – 12) = ms3 (28 – θ)
or s1 (θ -12) = s3 (28 − θ) = 16/15 s1 (28 – θ)
∴ θ = 628 / 31 = 20.3°C
— : end of Thermal Properties of Matter Numerical-II for ISC Class-11 Nootan Physics Ch-16 :–
Return to : – Nootan Solutions for ISC Class-11 Physics
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