In an experiment with Joule's apparatus, it was observed that the temperature of 0.5 kg of water rises by 1°C, when the mechanical energy transferred by falling weights is 2093 J. With the help of calculations, show that these values lead to the same conclusion as that of Joule ?
In this problem we have to calculate the mechanical energy required to raise the temperature of 1 kilogram of water by 1°C.
Here we have been given that:
Mechanical energy required to raise the temperature of 0.5 kg of water by 1°C = 2093 J
So,Mechanical energy required to raise the temperature of 1 kg of water by 1°C = 2093 × 1/0.5 × 1 = 4186 J
From these calculations we find that the mechanical energy required to raise the temperature of 1 kg of water by 1°C is 4186 joules. Now, this is the same conclusion as that of Joule because Joule also established by his experiments that 4186 joule of mechanical energy is required to raise the temperature of 1 kg of water by 1°C.
A body possessing 756 J of energy is made to fall in 600 g of water. If whole of the energy is converted into heat energy, calculate the rise in temperature of water? (Specific heat of water = 4200 J kg−1°C−1)
Since all of the 756 joules of energy possessed by the body is converted into heat energy, this means that 756 joules of heat energy will be available to heat 600 grams of water.
| Heat available, Q | = | 756 J |
| Mass of water , m | = | 600 g |
| = | 600/1000 | |
| = | 0.6 kg | |
| Specific heat of water, C | = | 4200 J kg − 1°C − 1 |
| And, Rise in temperature of water T | = | ? (To be calculated) |
| Now, putting the above values in the relation : | ||
| Q | = | m × C × T |
| We get, 756 | = | 0.6 × 4200 × T |
| So,T | = | 756/0.6 × 4200 |
| = | 0.3°C | |
| Thus, the rise in temperature of water is 0.3°C. |
A water – fall is 100 m high. How much is the rise in temperature of each kilogram of water when it strikes at the bottom of the fall ? (Specific heat of water = 4200 J/kg/°C; g = 10 m/s2?
The water at 100 m height has some potential energy which is to be transformed (converted) into heat energy. So, let us first calculate the potential energy contained in 1 kilogram of water at the 100 m height (because 'each kilogram' means '1 kilogram').
| Now,Potential energy | = | m × g × h |
| Here, Mass of water, m | = | 1 kg |
| Acceleration due to gravity, g | = | 10 m/s2 |
| Height, h | = | 100 m |
| So, Potential energy | = | 1 × 10 × 100 |
| = | 1000 joules ----- (1) |
Now, since there is an exact equivalence between mechanical energy and heat energy, so when 1000 joules of potential energy is lost, it will produce an equal amount, 1000 joules of heat energy. And this heat energy of 1000 joules is to be used for heating 1 kilogram of water.
| Now, Heat available, Q | = | 1000 J |
| Mass of water, m | = | 1 kg |
| Specific heat of water, C | = | 4200 J/kg/°C |
| Rise in temperature, T | = | ? (To be calculated) |
| We know that, Q | = | m × C × T |
| so 1000 | = | 1 × 4200 × T |
| T | = | 1000/1 × 4200 |
| = | 0.238°C |
| Thus, the rise in temperature of water will be 0.238°C. |