A copper cylinder is initially at 20.0 oC. At what temperature in oC will its volume be 0.39 % larger than it is at 20.0 oC? The linear coefficient of expansion of copper is 28.6x10-6¢XC-1
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To find the temperature at which the volume of the copper cylinder will be 0.39% larger than its initial volume, we can use the formula for the volume expansion of a solid:
ΔV = V₀ * α * ΔT
Where:
ΔV is the change in volume,
V₀ is the initial volume,
α is the linear coefficient of expansion, and
ΔT is the change in temperature.
In this case, we want to find the temperature at which the volume will be 0.39% larger than its initial volume, which means ΔV = 0.0039 * V₀.
Let's substitute these values into the formula and solve for ΔT:
0.0039 * V₀ = V₀ * α * ΔT
Dividing both sides of the equation by V₀:
0.0039 = α * ΔT
Now we isolate ΔT:
ΔT = 0.0039 / α
Substituting the given value for α (linear coefficient of expansion for copper):
ΔT = 0.0039 / 28.6x10^(-6) °C^(-1)
Calculating the value of ΔT:
ΔT ≈ 136.36 °C
To find the temperature at which the volume will be 0.39% larger than its initial volume, we add ΔT to the initial temperature of 20.0 °C:
Final temperature = 20.0 °C + 136.36 °C ≈ 156.36 °C
Therefore, the temperature at which the volume of the copper cylinder will be 0.39% larger than it is at 20.0 °C is approximately 156.36 °C.