Heat is added to a silver bar of mass 0.034 kg and length 0.28 m. The bar expands by an amount of 3.8×10-4 m. How much heat was added?
I need the answer and the equation please thanks!
no work to show?
deltaL= L *coefflinearexpansion*deltaTemp
but deltaTem= heatadded/(mass*coefheating)
solve for heat added. The coefficnets of heating, and coefficinets of silver can be looked up.
To solve this problem, we can use the equation for linear expansion:
ΔL = α * L * ΔT
Where:
ΔL is the change in length of the bar
α is the coefficient of linear expansion for the material (in this case, silver)
L is the original length of the bar
ΔT is the change in temperature
In this case, we are given the change in length (ΔL = 3.8×10-4 m) and the original length (L = 0.28 m). We need to find the change in temperature (ΔT) and then use it to calculate the heat added.
Rearranging the equation, we get:
ΔT = ΔL / (α * L)
To solve for ΔT, we need to know the coefficient of linear expansion (α) for silver. The coefficient of linear expansion for silver is approximately 0.000019 (1/°C).
Substituting the given values, we have:
ΔT = 3.8×10-4 m / (0.000019 /°C * 0.28 m)
ΔT = 3.8×10-4 / (0.000019 * 0.28 /°C)
Simplifying, we can find:
ΔT ≈ 6.42°C
Now, we can calculate the heat added using the equation:
Q = m * c * ΔT
Where:
Q is the heat added
m is the mass of the bar (0.034 kg)
c is the specific heat capacity of silver (approximately 235 J/kg°C)
Substituting the given values, we have:
Q = 0.034 kg * 235 J/kg°C * 6.42°C
Simplifying, we find:
Q ≈ 54.98 J
Therefore, approximately 54.98 Joules of heat were added to the silver bar.