A granite column in a building is 8.50 m high and has a mass of 2.16 multiplied by 104 kg.
(a) How much heat (J) is needed to increase its temperature by 10.5°C, given its specific heat is 0.200 kcal/kg·°C?
(b) What is its change in length (m) if its coefficient of linear expansion is 2.50 multiplied by 10-6/°C?
(c) Calculate the force (N) it would exert if not allowed to expand, given its cross-sectional area to be 1.00 m2 and the Young's modulus for granite to be 45.0 multiplied by 109 N/m2. Note that this force is much larger than the weight it supports, so that it will expand almost freely.
(d) How much work (J) does it perform when it expands, if it supports a mass of 1.00 multiplied by 104 kg?
(e) What is its change in internal energy(J)?
0.2kcal/kg•℃=837.36 J/kg•℃
m = ρV=837.36•
(a) Q=mcΔT=2.16•10⁴•837.36•10.5=1.9•10⁸ J
(b) α = 2.5•10⁻⁶ /℃
ΔL = α•L•ΔT = 2.5•10⁻⁶•8.5•10.5=2.23•10⁻⁴ m
(c) E= 45•10⁹ N/m²,
A=1 m²
σ=Eε
F/A = E•ΔL/L
F= A•E•ΔL/L =
=1•45•10⁹•2.23•10⁻⁴/8.5=1.18•10⁶ N
(d)
W= m₀•g•ΔL =1•10⁴•9.8•2.23•10⁻⁴=21.85 J
(e) ΔU = W+Q = 21.85+ 1.9•10⁸ ≈1.9•10⁸ J
(a) Well, that granite column is going to need some heat to warm up! To find out how much heat is needed to increase its temperature, we can use the formula:
Q = mcΔT
Where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Plugging in the values, we have:
Q = (2.16 x 10^4 kg) x (0.200 kcal/kg·°C) x (10.5°C)
Now, since we're using the metric system and not a mix of units, let's convert that equation and give you the answer in Joules:
1 kcal = 4184 J
Q = (2.16 x 10^4 kg) x (0.200 kcal/kg·°C) x (10.5°C) x (4184 J/kcal)
So, the amount of heat needed to increase its temperature is a whopping Q joules!
(b) To calculate the change in length, we can use the equation:
ΔL = αL₀ΔT
Where ΔL is the change in length, α is the coefficient of linear expansion, L₀ is the original length, and ΔT is the change in temperature.
Plugging in the values, we have:
ΔL = (2.50 x 10^-6/°C) x (8.50 m) x (10.5°C)
Oh, it seems like that granite column is going to experience a tiny change in length ΔL!
(c) Now, let's calculate the force that the granite column would exert if not allowed to expand. We can use the formula:
F = YAΔL/L₀
Where F is the force, Y is the Young's modulus, A is the cross-sectional area, ΔL is the change in length, and L₀ is the original length.
Let's plug in the values:
F = (45.0 x 10^9 N/m²) x (1.00 m²) x (ΔL) / (8.50 m)
Estimated force, prepare to behold the power of the granite column!
(d) To calculate the work performed when the granite column expands, we can use the formula:
W = FΔL
Where W is the work, F is the force, and ΔL is the change in length.
Let's calculate the work:
W = (F) x (ΔL)
Now, that's some magnificent work from the granite column!
(e) Woah, hang on a second! You want to know the change in internal energy? Well, let me tell you, the change in internal energy is zero. That's because the granite column is not subjected to any temperature change, compression, or stretching. It's just chilling and supporting weight like a boss!
And there you have it, all the answers to your granite column questions! Hope I was able to rock your world with my answers.
(a) The amount of heat required to increase the temperature of an object can be calculated using the formula:
Q = mcΔT
where:
Q = heat energy (in joules)
m = mass of the object (in kg)
c = specific heat capacity (in J/kg·°C)
ΔT = change in temperature (in °C)
Plugging in the values given in the question:
m = 2.16 × 10^4 kg
c = 0.200 kcal/kg·°C × 4186 J/kcal (since 1 kcal = 4186 J)
ΔT = 10.5°C
Q = (2.16 × 10^4 kg) × (0.200 kcal/kg·°C × 4186 J/kcal) × (10.5°C)
Q = 1.82256 × 10^7 J
Therefore, the amount of heat needed to increase the temperature of the granite column by 10.5°C is 1.82256 × 10^7 joules.
(b) The change in length of an object due to temperature can be calculated using the formula:
ΔL = αL₀ΔT
where:
ΔL = change in length (in meters)
α = coefficient of linear expansion (in 1/°C)
L₀ = initial length (in meters)
ΔT = change in temperature (in °C)
Plugging in the values given in the question:
α = 2.50 × 10^(-6)/°C
L₀ = 8.50 m
ΔT = 10.5°C
ΔL = (2.50 × 10^(-6)/°C) × (8.50 m) × (10.5°C)
ΔL = 2.16425 × 10^(-4) m
Therefore, the change in length of the granite column is 2.16425 × 10^(-4) meters.
(c) The force exerted by the column can be calculated using the formula:
F = YAΔL / L₀
where:
F = force (in newtons)
Y = Young's modulus (in N/m²)
A = cross-sectional area (in m²)
ΔL = change in length (in meters)
L₀ = initial length (in meters)
Plugging in the values given in the question:
Y = 45.0 × 10^9 N/m²
A = 1.00 m²
ΔL = 2.16425 × 10^(-4) m
L₀ = 8.50 m
F = (45.0 × 10^9 N/m²) × (1.00 m²) × (2.16425 × 10^(-4) m) / (8.50 m)
F = 1.276 × 10^6 N
Therefore, the force exerted by the granite column, if not allowed to expand, is 1.276 × 10^6 newtons.
(d) The work done by the granite column when it expands can be calculated using the formula:
W = F × ΔL
where:
W = work (in joules)
F = force (in newtons)
ΔL = change in length (in meters)
Plugging in the values given in the question:
F = 1.276 × 10^6 N
ΔL = 2.16425 × 10^(-4) m
W = (1.276 × 10^6 N) × (2.16425 × 10^(-4) m)
W = 275.9274 J
Therefore, the work performed by the granite column when it expands is 275.9274 joules.
(e) The change in internal energy can be calculated using the formula:
ΔU = Q - W
where:
ΔU = change in internal energy (in joules)
Q = heat energy (in joules)
W = work (in joules)
Plugging in the values calculated in parts (a) and (d):
Q = 1.82256 × 10^7 J
W = 275.9274 J
ΔU = 1.82256 × 10^7 J - 275.9274 J
ΔU = 1.81941 × 10^7 J
Therefore, the change in internal energy of the granite column is 1.81941 × 10^7 joules.
To solve these questions, we'll need to use the appropriate formulas and equations. Let's break down each question and explain how to find the answers.
(a) How much heat (J) is needed to increase its temperature by 10.5°C, given its specific heat is 0.200 kcal/kg·°C?
The formula to calculate heat (Q) is given as:
Q = mcΔT
Where:
Q = heat (in joules)
m = mass (in kilograms)
c = specific heat (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
To solve this question, we substitute the given values:
m = 2.16 x 10^4 kg
c = 0.200 kcal/kg·°C (1 kcal = 4184 J)
ΔT = 10.5°C
First, convert specific heat from kcal/kg·°C to J/kg·°C:
c = 0.200 kcal/kg·°C x 4184 J/kcal = 836.8 J/kg·°C
Now calculate the heat (Q):
Q = (2.16 x 10^4 kg) x (836.8 J/kg·°C) x (10.5°C)
Calculate the result.
(b) What is its change in length (m) if its coefficient of linear expansion is 2.50 x 10^-6/°C?
The formula to calculate the change in length (ΔL) is given as:
ΔL = αLΔT
Where:
ΔL = change in length
α = coefficient of linear expansion (in per degree Celsius)
L = original length
ΔT = change in temperature (in degrees Celsius)
To solve this question, we substitute the given values:
α = 2.50 x 10^-6/°C
L = 8.50 m
ΔT = 10.5°C
Now calculate the change in length (ΔL):
ΔL = (2.50 x 10^-6/°C) x (8.50 m) x (10.5°C)
Calculate the result.
(c) Calculate the force (N) it would exert if not allowed to expand, given its cross-sectional area to be 1.00 m² and the Young's modulus for granite to be 45.0 x 10^9 N/m².
The formula to calculate the force (F) due to the restriction of expansion is given as:
F = YΔL × A
Where:
F = force
Y = Young's modulus
ΔL = change in length
A = cross-sectional area
To solve this question, we substitute the given values:
Y = 45.0 x 10^9 N/m²
ΔL = (calculated in part (b))
A = 1.00 m²
Now calculate the force (F):
F = (45.0 x 10^9 N/m²) x (ΔL from part (b)) x (1.00 m²)
Calculate the result.
(d) How much work (J) does it perform when it expands, if it supports a mass of 1.00 x 10^4 kg?
The formula to calculate work (W) done by a force is given as:
W = Fs
Where:
W = work done (in joules)
F = force (from part (c))
s = displacement (change in length, ΔL, from part (b))
To solve this question, we substitute the given values:
F = (calculated in part (c))
s = ΔL (from part (b))
Now calculate the work (W):
W = (F from part (c)) x (ΔL from part (b))
Calculate the result.
(e) What is its change in internal energy (J)?
Since the granite column is not undergoing any phase change or experiencing any non-conservative forces, its change in internal energy (ΔU) is given by:
ΔU = Q - W
Where:
ΔU = change in internal energy
Q = heat (from part (a))
W = work (from part (d))
To solve this question, substitute the given values:
Q = (from part (a))
W = (from part (d))
Now calculate the change in internal energy (ΔU):
ΔU = (Q from part (a)) - (W from part (d))
Calculate the result.