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प्रश्न
Describe a method to determine the specific heat capacity of a solid (say, a piece of copper).
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उत्तर
First we weigh the given piece of solid and note its mass m1. Then it is heated by suspending it inside a heater. Then a known mass (say m2) of water is taken in a thin glass beaker and its temperature θ1 is recorded with a thermometer. When the given piece of solid becomes heated, its temperature θ2 is noted and it is quickly dropped into the water contained in the beaker, such that no water splashes out. The contents of beaker are well stirred and the final highest temperature θ3 is noted.
Assuming that the heat capacity of beaker is negligible and there is not heat loss to the surroundings,
Heat lost by the solid = Heat gained by water Mass of solid × Specific heat capacity of solid × Fall in temperature of solid = Mass of water × Sp. capacity of water × Rise in temperature of water.
or m1 × c × (θ2 - θ3) = m3 × 4.2 × (θ3 - θ1)
or c = `("m"_2 xx 4.2 xx (theta_3 - theta_1))/(theta_1 xx (theta_2 - theta_3))` J/g°C
Here we have assumed that specific heat capacity of water is 4.2 J/g°C.
संबंधित प्रश्न
What do you understand by the following statements:
The heat capacity of the body is 60JK-1.
What do you understand by the following statements:
The specific heat capacity of lead is 130 Jkg-1K-1.
(b) 2000 J of heat energy is required to raise the temperature of 4 kg of a
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Water boils at 120 °C in a pressure cooker. Explain the reason.
What is heat? What is the S. I. unit of heat?
Write the approximate values of the specific latent heat of vaporization of steam.
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Two metals A and B have specific heat capacities in the ratio 2:3. If they are supplied same amount of heat then
Which metal piece will have greater mass if the rise in temperature is the same for both metals?
A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The xg of ice melts as the temperature of the water reaches 0°C. The value of x is ______.
[Use: Specific heat capacity of water = 4200 Jkg-1K-1, Latent heat of ice = 3.5 × 105 Jkg-1]
