A23.9 g sample of iridium is heated to 89.7°c, and then dropped into 20.0 g of water in a foam-cup calorimeter. the temperature of the water went from 20.1°c to 22.6°c. calculate the specific heat of iridium (specific heat of water = 4.18 j/g.°c)

The specific heat of iridium is 0.130 J/g°C

Explanation:

When iridium is dipped in water, the amount of heat released by iridium will be equal to the amount of heat absorbed by water.

The equation used to calculate heat released or absorbed follows:

     ......(1)

where,

q = heat absorbed or released

= mass of iridium = 23.9 g

= mass of water = 20.0 g

= final temperature = 22.6°C

= initial temperature of iridium = 89.7°C

= initial temperature of water = 20.1°C

= specific heat of iridium = ?

= specific heat of water = 4.18 J/g°C

Putting values in equation 1, we get:

Hence, the specific heat of iridium is 0.130 J/g°C

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