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

Answer:

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

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

Answer 2

Answer:

0.13J/g°C

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 $C_{I}$;

$C_{I}$ = $\frac{209}{1603.69}$

$C_{I}$ = 0.13

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