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प्रश्न
An electric immersion heater is rated 1250 W. Calculate the time in which it will heat 20 kg of water at 5°C to 65°C.
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उत्तर
Power supplied by an electric heater (VI) = 1250 W
θ1 = Initial temperature = 5°C
m = Mass of water = 20 kg
θ2 = Final temperature = 65°C
c = specific heat capacity by heater = Heat gained by water ...(Law of calorimeter)
VIt = mc (θ2 - θ1) .....(Q = VI t)
1250 t = 20 × 4200 × 60 ....[(65°C - 5°C = 60°C)]
1250 t = 5040000
t = 40325 seconds
संबंधित प्रश्न
A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.
Define heat capacity and state its SI unit.
Heat energy is supplied at a constant rate to 100g of ice at 0 °C. The ice is converted into water at 0° C in 2 minutes. How much time will be required to raise the temperature of water from 0 °C to 20 °C? [Given: sp. heat capacity of water = 4.2 J g-1 °C-1, sp. latent heat of ice = 336 J g-1].
63.2 g of copper at 50°C can just melt 3.8g of ice. If the specific latent heat of ice is 336 J/g, find the specific heat capacity of copper.
Derive an expression for finding out the specific heat capacity of a body (solid) from the readings of an experiment given below:
(i) Mass of empty calorimeter (with stirrer) = m1 gm
(ii) Mass of the metal piece = M gm
(iii) Mass of colorimeter and water = m2 gm
(iv) Initial temperature and water = t1°C
(v) Temperature of hot solid (metal piece) = t2 °C
(vi) Final temperature of the mixture = t°C
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Read the passage and answer the questions based on it.
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy. The change in temperature continues till the temperatures of both objects attain the same value. In this process, the cold object gains heat energy and the hot object loses heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat-resistant box then no energy can flow from inside the box or come into the box. In this situation, we get the following principle.
Heat energy lost by the hot object = Heat energy gained by the cold object. This is called the ‘Principle of heat exchange’.
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Which metal piece will show a greater rise in temperature given their masses is the same?
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Which substance will have more specific heat capacity?
