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A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by :(Given: room temperature = 30°C, the specific heat of copper = 0.1cal/gm°C)


A

1250°C

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B

825°C

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C

800°C

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D

885°C

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Solution

The correct option is D

885°C


The explanation for the correct answer:-

Option (D) 885°C:

Step 1: Given data

Mass of copper ball= 100gm

Room temperature= 30°C

Mass of water= 170gm

Specific heat of copper = 0.1calgm°C-1

Temperature of system= 75°C

Step 2: Heat absorbed by the bodies

We know that heat lost or taken by a body for a change in its temperature ΔT is given as Q=mcΔT.

Where m= mass of the body and c= specific heat of the body

As a result, when the copper ball comes in contact with the copper calorimeter and the water which is inside, it loses heat. This heat will now be absorbed by the calorimeter and the water and its temperature will increase. It is also given that the equilibrium temperature of the three bodies is 75°C, and we know that the specific heat of water is 1calgm°C-1

heat lost by the copper ball is

Qb=mbcb(T-75°C)Qb=100×0.1×(T75)Qb=10(T75)

heat lost by the calorimeter is

Qc=mccc(7530)Qc=100×0.1×45Qc=450cal

.heat lost by the calorimeter is

Qw=mwcw(7530)Qw=170×1×45Qw=7650cal

Step 3: Calculating the temperature

We know that, Qb=Qc+Qw

10(T75)=450+765010T750=8100T=885010T=885°C

Thus the temperature is T=885°C

Therefore, T is given by T=885°C, option (D) is the correct answer.


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