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

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !