Question

A 23.9 g sample of iridium is heated to 89.7 oC, and then dropped into 20.0 g of water in a coffee-cup calorimeter. The temperature of the water went from 20.1 oC to 22.6 oC. Calculate the specific heat of iridium. (specific heat of water = 4.18 J/g oC).

Answer 1

0.13J/g°C

Step-by-step explanation:

The mixture of the Iridium in water is in a thermo-equilibrium since no heat is lost to the environment. i.e The heat lost (-
`H_(I)`) by Iridium is equal to the heat gained (
`H_(W)`) by water. This can be represented as follows;

-
`H_(I)` =
`H_(W)` --------------------------(i)

The negative sign shows that heat is lost to the environment...

But;

`H_(I)` =
`m_(I)`
`C_(I)` Δ
`T_(I)` --------------------------(ii)

Where;

`m_(I)` = mass of Iridium

`C_(I)` = specific heat capacity of Iridium

Δ
`T_(I)` = change in temperature of Iridium =
`T_(I2)` -
`T_(I1)`

`T_(I2)` = final temperature of Iridium

`T_(I1)` = initial temperature of Iridium

Also;

`H_(W)` =
`m_(W)`
`C_(W)` Δ
`T_(W)` ------------------------(iii)

Where;

`m_(W)` = mass of water

`C_(W)` = specific heat capacity of water

Δ
`T_(W)` = change in temperature of water =
`T_(W2)` -
`T_(W1)`

`T_(W2)` = final temperature of water

`T_(W1)` = initial temperature of water

From the question;

`m_(I)` = 23.9g

`C_(I)` = ?

`T_(I2)` = 22.6°C [the same as the final temperature of water]

`T_(I1)` = 89.7°C

Δ
`T_(I)` = 22.6°C - 89.7°C = -67.1°C

`m_(W)` = 20.0g

`C_(W)` = 4.18 J/g °C

`T_(W2)` = 22.6°C

`T_(W1)` = 20.1°C

Δ
`T_(W)` = 22.6°C - 20.1°C = 2.5°C

Substitute the values of
`H_(W)` and
`H_(W)` into equation (i)

-
`m_(I)`
`C_(I)` Δ
`T_(I)` =
`m_(W)`
`C_(W)` Δ
`T_(W)` -------------------------------(iv)

Now substitute the values of all the variables in equation(iv) into the same;

- 23.9 x
`C_(I)` x - 67.1 = 20.0 x 4.18 x 2.5

1603.69
`C_(I)` = 209

Then, solve for
`C_(I)`;

`C_(I)` =
`(209)/(1603.69)`

`C_(I)` = 0.13

Therefore, the specific heat of Iridium is 0.13J/g°C

Answer 2

The specific heat capacity of iridium = 0.130 J/g°C

Step-by-step explanation:

Assuming no heat losses to the environment and to the calorimeter,

Heat lost by the iridium sample = Heat gained by water

Heat lost by the iridium sample = mC ΔT

m = mass of iridium = 23.9 g

C = specific heat capacity of the iridium = ?

ΔT = change in temperature of the iridium = 89.7 - 22.6 = 67.1°C

Heat lost by the iridium sample = (23.9)(C)(67.1) = (1603.69 C) J

Heat gained by water = mC ΔT

m = mass of water = 20.0 g

C = 4.18 J/g°C

ΔT = 22.6 - 20.1 = 2.5°C

Heat gained by water = 20 × 4.18 × 2.5 = 209 J

Heat lost by the iridium sample = Heat gained by water

1603.69C = 209

C = (209/1603.69) = 0.130 J/g°C