25 g of ice at – 20°C is changed into superheated steam at 140°C by supplying 78400 J of heat energy. If sp. heat capacity of ice and steam 2 Jg – 1°C – 1 the sp. latent heat of ice is 336 Jg – 1. Calculate the sp. latent heat capacity of vaporization of steam ?
| Heat absorbed by ice to attain temp. of 0°C | = | mcθR |
| = | 25 × 2 × 20 | |
| = | 1000 J | |
| Heat absorbed by ice to form water at 0°C | = | mLice |
| = | 25 × 336 | |
| = | 8400 J | |
| Heat absorbed by water to attain 100°C | = | mc θR |
| = | 25 × 4.2 × 100 | |
| = | 10500J | |
| Heat absorbed by water, to change to steam | = | mL |
| = | 25 g × L | |
| Heat absorbed by steam to attain 140°C | = | m × c × θR |
| = | 25 × 2 × 40 | |
| = | 2000 J | |
| ∴ Total heat absorbed | = | (1000 + 8400 + 10500 + 2000) J + 20g × L |
| = | 21900J + 25g × L | |
| 21900 J + 25 g × L | = | 78400 s |
| ∴L | = | 56500 / 25 |
| = | 2260 Jg – 1 |
Calculate the mass of steam which should be passed through 65 g of water, contained in a copper calorimeter of mass 42 g (S.H.C. 0.4 Jg – 1°C – 1) at 10°C, such that final temperature is 40°C. Take sp. latent heat of steam as 2240 Jg – 1 ?
| Substance | Mass | S.H.C/S.L.H | Initial Temp | Final Temp = 40°C |
|---|---|---|---|---|
| Steam | (m) | 2240 Jg – 1 | 100°C | θR = (40 – 10) = 30°C |
| Water cold | 65 g | 4.2 J/g °C | 10°C | θF = (100 – 40) = 60°C |
| Calorimeter | 42 g | 0.4 J/g °C | 10°C | θF = (100 – 40) = 60°C |
| Heat given by steam to form water at 100°C | = | mL |
| = | m × 2240 Jg – 1 | |
| Heat given out by water at 100°C | = | mc θF |
| = | m × 4.2 × 60 | |
| = | 252 m Jg – 1 | |
| ∴Total heat given out | = | (2240 + 252) m |
| = | 2492 m Jg – 1 | |
| Heat gained by water (cold) | = | mc θR |
| = | 65 × 4.2 × 30 | |
| = | 8190 J | |
| Heat gained by calorimeter | = | mc θR |
| = | 42 × 0.4 × 30 | |
| = | 504 J | |
| Total heat given out | = | (8190 + 504) |
| = | 8694 J | |
| Heat lost | = | Heat gained |
| 2492 m Jg – 1 | = | 8694 J |
| ∴ m | = | 8694 / 2492 |
| = | 3.489 g. |
Steam at 100°C is passed through a copper vessel of mass 820 g at 20°C, till the steam stops condensing in vessel. If 11.70 g of water is condensed in the vessel, calculate the sp. latent heat of vaporization of steam. Specific heat capacity of copper is 0.4 Jg – 1°C – 1 ?
When the steam stops condensing, the final temp. is 100°C.
| Substance | Mass | S.H.C/S.L.H | Initial Temp | Final Temp = 100°C |
|---|---|---|---|---|
| Copper vessel | 820 g | 0.4 J/g °C | 20°C | θR = (100 – 20) = 80°C |
| Steam | 11.70g | (L) | 100°C | θF = (100 – 100) = 0°C |
| Heat gained by copper vessel | = | mc θR |
| = | 820 × 0.4 × 80 | |
| = | 26240 J | |
| Heat lost by steam to form water at 100°C | = | mL |
| = | 11.70 g × L | |
| Heat lost | = | Heat gained |
| 11.70 g × L | = | 26240 |
| L | = | 2242.74 Jg – 1 |
A beaker contains 1450 g of ice. Into this beaker is passed steam till 200 g of ice melts. The amount of water formed is found to be 225 g. Calculate the specific latent heat of vaporization of steam. Specific latent heat of fusion of ice is 336 Jg – 1?
| Amount of ice melted | = | 200 g |
| ∴ Amount of steam condensed | = | (225 – 200) |
| = | 25 g |
| Substance | Mass | S.H.C/S.L.H | Initial Temp | Final Temp = 0°C |
|---|---|---|---|---|
| Ice | 200 g | 336 Jg – 1 | 0°C | ∴ θR = 0°C |
| Steam | 25 g | (L) | 100° | ∴ θF = 100°C |
| Heat gained by ice | = | mL |
| = | 200 × 336 | |
| = | 67200 J | |
| Heat lost by steam to form water at 100°C | = | mL |
| = | 25 g × L | |
| Heat lost by water at 100°C | = | mc θF |
| = | 25 × 4.2 × 100 | |
| = | 10500 J | |
| Heat lost | = | Heat gained |
| 25 g × L + 10500 J | = | 67200 |
| ∴ L | = | 56700 / 25 |
| = | 2268 Jg – 1 |
A brass rod of 0.2 kg mass at 100°C is dropped into 0.5 kg of water at 20°C. The final temperature is 23°C. Calculate the specific heat of brass (Specific heat of water = 4.18 × 103 J/kg/°C) ?
| (i) Calculation of Heat Lost by Brass Rod | ||
| Mass of brass rod, m | = | 0.2 kg |
| Specific heat of brass, C1 | = | ? (To be calculated) |
| Initial temperature of brass rod | = | 100°C |
| Final temperature of brass rod | = | 23°C |
| So,Change (fall) in temp, of brass rod, T | = | 100° – 23° |
| = | 77°C | |
| Thus, Heat lost by brass rod, Q1 | = | m1 × C1 × T1 |
| = | 0.2 × C1 × 77 | |
| = | 15.4 × C1 joules ------(i) | |
| (ii) Calculation of Heat Gained by Water | ||
| Mass of water, m2 | = | 0.5 kg |
| Specific heat of water, C2 | = | 4.18 × 103 J/kg/°C |
| Initial temperature of water | = | 20°C |
| Final temperature of water | = | 23°C |
| So,Change (rise) in temp, of water, T2 | = | 23° – 20° |
| = | 3°C | |
| Thus, Heat gained by water, Q2 | = | m2 × C2 × T2 |
| = | 0.5 × 4.18 × 103 × 3 | |
| = | 6270 joules ------ (2) | |
| Now, according to the principle of calorimetry : | ||
| Heat lost | = | Heat gained |
| So, 15.4 × C1 | = | 6270 |
| And, C1 | = | 6270 / 15.4 |
| = | 407 J/kg/°C | |
| Thus, the specific heat of brass is 407 J/kg/°C. |