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Answer 1

Answer:

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Answer 2

Question:

A calorimeter of water equivalent 100 grams  contains 200 grams of water at 10°C. A solid of  mass 500 grams at 45° C is added to the  calorimeter. If equilibrium temperature is 25°C,  then the specific heat of the solid is

Answer:

The specific heat of the solid is 0.45 cal/g-°C

Explanation:

>> A calorimeter of water equivalent 100 grams  contains 200 grams of water at 10°C.

The total mass of water equivalent and water,

  m = 100 g + 200 g = 300 g

>> The equilibrium temperature is 25°C  

  The change in temperature,

 ∆T = 25°C - 10°C = 15°C

>> The specific heat of water, s = 1 cal/g-°C

We know, Q = ms∆T

The heat gained by water

 = 300 g × 1 cal/g-°C × 15°C  

= 4500 cal

____________________________

SOLID :

Let the specific heat of the solid be S cal/g-°C

>> The mass of solid, M = 500 g

>> The equilibrium temperature is 25°C

So, the change in temperature,

 ∆Tₛ = 45°C - 25°C = 20°C

 

The heat lost by solid = MS∆Tₛ  

= 500 g × S cal/g-°C  × 20°C

= 10000S cal

Heat gained by water = Heat lost by solid

4500 = 10000S

S = 4500/10000

S = 0.45 cal/g-°C  

Therefore, the specific heat of the solid is 0.45 cal/g-°C