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Hint: In question when rate of cooling is to be calculated then rate of loss of heat or rate of cooling is always taken as directly proportional to the product of mass and specific heat of the material. That material which has the highest value of this product will have the highest rate of loss of heat or rate of cooling.
Complete step-by-step solution:
According to Newton’s Law of Cooling “Rate of loss of heat is directly proportional to temperature difference between system and surrounding”
Let us assume rate of loss of heat can be expressed as \[\dfrac{dQ}{dt}\]
Let us assume the temperature of the system is T and the temperature of the surrounding is T0.
\[-\dfrac{dQ}{dt}\alpha (T-{{T}_{0}})\]
\[\Rightarrow \dfrac{dQ}{dt}=-k(T-{{T}_{o}})\]
Where k is a constant.
Since we know that,
\[dQ=mcdT\]
Divide by dt on both sides we get ,
\[\dfrac{dQ}{dt}=mc\dfrac{dT}{dt}\] - - - - - - - - Equation 1
Here c represents the specific heat
This equation represents the rate of loss of heat.
In question it is given that four spheres A, B, C, D are of different material but their radius is the same.
Ratio of their densities is given: - 2:3:5:1
Ratio of their specific heat is given: -3:6:2:4
since it is given in question that four spheres are kept at the same temperature so \[\dfrac{dT}{dt}\]is the same for all spheres.
So equation 1 becomes
\[\dfrac{dQ}{dt}\alpha m\times c\] - - - - - - - - Equation 2
since we know that
\[mass=density\times volume\]
Let us assume density be d and volume be V
So equation 2 becomes,
\[\dfrac{dQ}{dt}\alpha d\times V\times c\].
Since radius is same for all spheres so this equation can be written as,
\[\dfrac{dQ}{dt}\alpha d\times c\]
Rate of loss of heat is directly proportional to the product of density and specific heat, the body which has the highest value of product of density and specific heat will have the highest rate of loss of heat or highest rate of cooling.
Sphere A has density value =2 and specific heat value =3
\[\dfrac{dQ}{dt}=2\times 3=6\]
Sphere B has density value =3 and specific heat value =6
\[\dfrac{dQ}{dt}=3\times 6=18\]
Sphere C has density value =5 and specific heat value =2
\[\dfrac{dQ}{dt}=5\times 2=10\]
Sphere D has density value =1 and specific heat value =4
\[\dfrac{dQ}{dt}=1\times 4=4\]
Since sphere B has the highest value of rate of loss of heat. So we can conclude that sphere B has the highest rate of cooling.
Correct option is C.
Note:When heat is given to any substance then it either changes only the state of the substance or it will change the temperature of the substance as well as state. When only the state of the substance is changed then the concept of Latent Heat is used and when both temperature and state of substance is changing then the concept of specific heat is used.
Complete step-by-step solution:
According to Newton’s Law of Cooling “Rate of loss of heat is directly proportional to temperature difference between system and surrounding”
Let us assume rate of loss of heat can be expressed as \[\dfrac{dQ}{dt}\]
Let us assume the temperature of the system is T and the temperature of the surrounding is T0.
\[-\dfrac{dQ}{dt}\alpha (T-{{T}_{0}})\]
\[\Rightarrow \dfrac{dQ}{dt}=-k(T-{{T}_{o}})\]
Where k is a constant.
Since we know that,
\[dQ=mcdT\]
Divide by dt on both sides we get ,
\[\dfrac{dQ}{dt}=mc\dfrac{dT}{dt}\] - - - - - - - - Equation 1
Here c represents the specific heat
This equation represents the rate of loss of heat.
In question it is given that four spheres A, B, C, D are of different material but their radius is the same.
Ratio of their densities is given: - 2:3:5:1
Ratio of their specific heat is given: -3:6:2:4
since it is given in question that four spheres are kept at the same temperature so \[\dfrac{dT}{dt}\]is the same for all spheres.
So equation 1 becomes
\[\dfrac{dQ}{dt}\alpha m\times c\] - - - - - - - - Equation 2
since we know that
\[mass=density\times volume\]
Let us assume density be d and volume be V
So equation 2 becomes,
\[\dfrac{dQ}{dt}\alpha d\times V\times c\].
Since radius is same for all spheres so this equation can be written as,
\[\dfrac{dQ}{dt}\alpha d\times c\]
Rate of loss of heat is directly proportional to the product of density and specific heat, the body which has the highest value of product of density and specific heat will have the highest rate of loss of heat or highest rate of cooling.
Sphere A has density value =2 and specific heat value =3
\[\dfrac{dQ}{dt}=2\times 3=6\]
Sphere B has density value =3 and specific heat value =6
\[\dfrac{dQ}{dt}=3\times 6=18\]
Sphere C has density value =5 and specific heat value =2
\[\dfrac{dQ}{dt}=5\times 2=10\]
Sphere D has density value =1 and specific heat value =4
\[\dfrac{dQ}{dt}=1\times 4=4\]
Since sphere B has the highest value of rate of loss of heat. So we can conclude that sphere B has the highest rate of cooling.
Correct option is C.
Note:When heat is given to any substance then it either changes only the state of the substance or it will change the temperature of the substance as well as state. When only the state of the substance is changed then the concept of Latent Heat is used and when both temperature and state of substance is changing then the concept of specific heat is used.
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