
If the specific heat capacity of metal is $ 200\text{ }J\text{ }k{{g}^{-1}}{{K}^{-1}}, $ how much additional heat energy will the metal give out on cooling to $ -50{}^\circ C $ ? $ \left[ \Delta t=850 \right] $
(A) $ 25,000\text{ }J $
(B) $ 25,500\text{ }J $
(C) $ 24,500\text{ }J $
(D) None of the above.
Answer
412.2k+ views
Hint :We know the principle of the calorimeter, which also explains the law of conservation of energy. Also use the formula for the energy or heat exchanged in terms of mass of the substance, specific heat of the substance and change in temperature of the substance.
Complete Step By Step Answer:
As we know that the thermal capacity is defined as the property of a material to absorb heat when the material is heated and to release heat when the material is cooled down. We also know that, according to the law of conservation of energy, energy can neither be created nor be destroyed; rather it can only be transferred from one form to another.
Further, according to the principle of calorimeter the total heat lost by a hot body or material is equal to the total heat gained by the cold body or material. It can be written as Heat Lost is equal to Heat Gained. This principle is on the basis of the law of conservation of energy. As we know that energy is defined as the ability to do work. Energy can be found in many things and can take different forms, like-kinetic energy is the energy of motion, and potential energy is energy due to an object's position or structure. Total energy of a system is given by the sum of kinetic energy and potential energy.
Heat is the degree of hotness or coldness of the body. It is calculated by given formula
$ Q=ms\Delta T, $ where $ \,m=0.15= $ Mass of Body which is predefined value,
$ S=200= $ Specific Heat Capacity,
$ \Delta T=\left( {{T}_{2}}-{{T}_{1}} \right)=850 $ Change in Temperature and
$ Q= $ Amount of Heat.
Amount of heat given out $ \left( Q \right)=ms\left( {{T}_{2}}-{{T}_{1}} \right)\text{ }=0.15\times 200\times \left( 850 \right) $
$ \Rightarrow Q=25500\text{ }J $
Therefore, the correct answer is option B.
Note :
Remember that the laws of conservation of energy can never be violated in any case. According to the law of conservation of energy, energy can neither be created nor be destroyed; rather it can only be transferred from one form to another.
Complete Step By Step Answer:
As we know that the thermal capacity is defined as the property of a material to absorb heat when the material is heated and to release heat when the material is cooled down. We also know that, according to the law of conservation of energy, energy can neither be created nor be destroyed; rather it can only be transferred from one form to another.
Further, according to the principle of calorimeter the total heat lost by a hot body or material is equal to the total heat gained by the cold body or material. It can be written as Heat Lost is equal to Heat Gained. This principle is on the basis of the law of conservation of energy. As we know that energy is defined as the ability to do work. Energy can be found in many things and can take different forms, like-kinetic energy is the energy of motion, and potential energy is energy due to an object's position or structure. Total energy of a system is given by the sum of kinetic energy and potential energy.
Heat is the degree of hotness or coldness of the body. It is calculated by given formula
$ Q=ms\Delta T, $ where $ \,m=0.15= $ Mass of Body which is predefined value,
$ S=200= $ Specific Heat Capacity,
$ \Delta T=\left( {{T}_{2}}-{{T}_{1}} \right)=850 $ Change in Temperature and
$ Q= $ Amount of Heat.
Amount of heat given out $ \left( Q \right)=ms\left( {{T}_{2}}-{{T}_{1}} \right)\text{ }=0.15\times 200\times \left( 850 \right) $
$ \Rightarrow Q=25500\text{ }J $
Therefore, the correct answer is option B.
Note :
Remember that the laws of conservation of energy can never be violated in any case. According to the law of conservation of energy, energy can neither be created nor be destroyed; rather it can only be transferred from one form to another.
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