A copper kettle contains water at 29.1 °C. When the water is heated to 86.6 °C, the volume of the kettle expands by 5.56 x 10-5 m3. Determine the volume of the kettle at 29.1 °C.
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To determine the volume of the kettle at 29.1 °C, we need to use the principle of thermal expansion. The principle states that when a material is heated, it expands in volume.
We can use the formula for thermal expansion:
ΔV = V_0 * β * ΔT
where:
- ΔV is the change in volume
- V_0 is the initial volume
- β is the coefficient of thermal expansion
- ΔT is the change in temperature
In this case, we know the change in volume (ΔV = 5.56 x 10^(-5) m^3), the initial temperature (T_0 = 29.1 °C), and the final temperature (T_final = 86.6 °C).
To find the coefficient of thermal expansion (β), we need to look up the value for copper. The coefficient of thermal expansion for copper is approximately 1.7 x 10^(-5) °C^(-1).
Now, we can rearrange the formula to solve for the initial volume (V_0):
V_0 = ΔV / (β * ΔT)
Substituting in the given values:
V_0 = (5.56 x 10^(-5) m^3) / (1.7 x 10^(-5) °C^(-1) * (86.6 °C - 29.1 °C))
Simplifying the equation:
V_0 = (5.56 x 10^(-5) m^3) / (1.7 x 10^(-5) °C^(-1) * 57.5 °C)
V_0 ≈ 0.00380 m^3
Therefore, the volume of the kettle at 29.1 °C is approximately 0.00380 m^3.