Ametal object with a mass of 19g is heated to 96c, then transferred to a calorimeter containg 75g of water at 18c. the water and metal object reach a final temperature of 22c. a. what is the specific heat of this metal object?

Let us say that Cp is the specific heat of the metal object. Then we do a heat balance (heat lost by metal = heat gained by water):

- 19g * Cp * (22degC – 96degC) = 75g * 4.184J/g degC * (22degC – 18degC)

Cp = 0.893 J/g degC

Answer;

Specific heat of the metal object  = 0.893 J/g°C  

Explanation and solution;

Quantity of heat = mass × specific heat capacity × change in temperature

ΔQ = Mcθ

Heat lost by the metal object = Heat gained by water

Heat lost by the metal object

= Mass of the metal object × specific heat of the metal ×change in temp

= 19 x specific heat ×( 22 - 96)

= 1406 × specific heat

Heat gained by water

= Mass of water × specific heat capacity of water × change in temperature

= 75 x 4.186 x ( 22 - 18)  

= 1255.8  J

Therefore;

= 1406 x specific heat = 1255.8  

Specific heat = 0.893 J/g°C  

that would be because of both gravity and friction. so     c

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