Respuesta :
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:
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 (-[tex]H_{I}[/tex]) by Iridium is equal to the heat gained ([tex]H_{W}[/tex]) by water. This can be represented as follows;
- [tex]H_{I}[/tex] = [tex]H_{W}[/tex] --------------------------(i)
The negative sign shows that heat is lost to the environment...
But;
[tex]H_{I}[/tex] = [tex]m_{I}[/tex] [tex]C_{I}[/tex] Δ[tex]T_{I}[/tex] --------------------------(ii)
Where;
[tex]m_{I}[/tex] = mass of Iridium
[tex]C_{I}[/tex] = specific heat capacity of Iridium
Δ[tex]T_{I}[/tex] = change in temperature of Iridium = [tex]T_{I2}[/tex] - [tex]T_{I1}[/tex]
[tex]T_{I2}[/tex] = final temperature of Iridium
[tex]T_{I1}[/tex] = initial temperature of Iridium
Also;
[tex]H_{W}[/tex] = [tex]m_{W}[/tex] [tex]C_{W}[/tex] Δ[tex]T_{W}[/tex] ------------------------(iii)
Where;
[tex]m_{W}[/tex] = mass of water
[tex]C_{W}[/tex] = specific heat capacity of water
Δ[tex]T_{W}[/tex] = change in temperature of water = [tex]T_{W2}[/tex] - [tex]T_{W1}[/tex]
[tex]T_{W2}[/tex] = final temperature of water
[tex]T_{W1}[/tex] = initial temperature of water
From the question;
[tex]m_{I}[/tex] = 23.9g
[tex]C_{I}[/tex] = ?
[tex]T_{I2}[/tex] = 22.6°C [the same as the final temperature of water]
[tex]T_{I1}[/tex] = 89.7°C
Δ[tex]T_{I}[/tex] = 22.6°C - 89.7°C = -67.1°C
[tex]m_{W}[/tex] = 20.0g
[tex]C_{W}[/tex] = 4.18 J/g °C
[tex]T_{W2}[/tex] = 22.6°C
[tex]T_{W1}[/tex] = 20.1°C
Δ[tex]T_{W}[/tex] = 22.6°C - 20.1°C = 2.5°C
Substitute the values of [tex]H_{W}[/tex] and [tex]H_{W}[/tex] into equation (i)
- [tex]m_{I}[/tex] [tex]C_{I}[/tex] Δ[tex]T_{I}[/tex] = [tex]m_{W}[/tex] [tex]C_{W}[/tex] Δ[tex]T_{W}[/tex] -------------------------------(iv)
Now substitute the values of all the variables in equation(iv) into the same;
- 23.9 x [tex]C_{I}[/tex] x - 67.1 = 20.0 x 4.18 x 2.5
1603.69[tex]C_{I}[/tex] = 209
Then, solve for [tex]C_{I}[/tex];
[tex]C_{I}[/tex] = [tex]\frac{209}{1603.69}[/tex]
[tex]C_{I}[/tex] = 0.13
Therefore, the specific heat of Iridium is 0.13J/g°C
