Respuesta :
Answer:
The specific heat of the alloy [tex]C_{a} = 0.37 \frac{KJ}{Kg K}[/tex]
Explanation:
Mass of an alloy [tex]m_{a}[/tex] = 25 gm
Initial temperature [tex]T_{a}[/tex] = 100°c = 373 K
Mass of water [tex]m_{w}[/tex] = 90 gm
Initial temperature of water [tex]T_{w}[/tex] = 25.32 °c = 298.32 K
Final temperature [tex]T_{f}[/tex] = 27.18 °c = 300.18 K
From energy balance equation
Heat lost by alloy = Heat gain by water
[tex]m_{a}[/tex] [tex]C_{a}[/tex] [[tex]T_{a}[/tex] - [tex]T_{f}[/tex]] = [tex]m_{w}[/tex] [tex]C_w[/tex] ([tex]T_{f}[/tex] -[tex]T_{w}[/tex] )
25 × [tex]C_{a}[/tex] × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)
[tex]C_{a} = 0.37 \frac{KJ}{Kg K}[/tex]
This is the specific heat of the alloy.
The specific heat capacity of the alloy is 0.385 J/gºC
We'll begin by calculating the heat absorbed by the water. This can be obtained as follow:
Mass of water (M) = 90 g
Initial temperature of water (T₁) = 25.32 °C
Final temperature (T₂) = 27.18 °C
Change in temperature (ΔT) = T₂ – T₁ = 27.18 – 25.32 = 1.86 °C
Specific heat capacity of water (C) = 4.184 J/gºC
Heat absorbed (Q) =?
Q = MCΔT
Q = 90 × 4.184 × 1.86
Q = 700.4016 J
Thus, the heat absorbed by the water is 700.4016 J
- Finally, we shall determine the specific heat capacity of the alloy
Heat absorbed = Heat released
Heat absorbed = 700.4016 J
Heat released = –700.4016 J
Mass of alloy (M) = 25 g
Initial temperature of alloy (T₁) = 100 °C
Final temperature (T₂) = 27.18 °C
Change in temperature (ΔT) = T₂ – T₁ = 27.18 – 100 = –72.82 °C
Specific heat capacity of alloy (C) =?
Q = MCΔT
–700.4016 = 25 × C × –72.82
–700.4016 = –1820.5 × C
Divide both side by –1820.5
C = –700.4016 / –1820.5
C = 0.385 J/gºC
Therefore, the specific heat capacity of the alloy is 0.385 J/gºC
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