(Q1)determine the quantity of heatconducted in 30min through an iron plate2.0cm thick and 0.10m2 in area if the temperature of the two sides are 0c and 20c.the coefficent of thermal conductivity is 50.4j/smc (a)91000000j(b)23000000j(c)41000000j(d)33000000j (Q2) one end of a 30-cm long aluminum rod is exposed to a temperature of 500c while the other is maintained at 20c.the rod has the diameter of 2.5cm. if heat is conducted through the rod at the rate of 164.9j/s. calculate the thermal conductivity of Aluminum(a)311.6w/mc(b)209.9w/mc(c)506.3w/mc(d)457.1w/mc
Q1
ΔQ/Δt =- κ•A•ΔT/L = >
ΔQ =- κ•A•ΔTt/L=
=- 50.4•0.1•(-20) •30•60/0.02 =
= 9072000 ≈9100000 J
Q2
q= ΔQ/Δt =- κ•A•ΔT/L
κ = - q•L/ A•ΔT =
= - 4•q•L/ π•d²•ΔT=
=- 4• 164.9•0.3/3.14•(0.025)² •(20-500) =
=209.9 W/m•℃
(Q1) To determine the quantity of heat conducted through the iron plate, we can use the formula:
Q = k * A * ΔT / d * t
Where:
Q = Quantity of heat conducted
k = Coefficient of thermal conductivity
A = Area
ΔT = Temperature difference
d = Thickness
t = Time
Given information:
k = 50.4 J/(s*m*C)
A = 0.10 m^2
ΔT = 20°C - 0°C = 20°C
d = 2.0 cm = 0.02 m
t = 30 min = 30 * 60 s = 1800 s
Substituting the values into the formula:
Q = 50.4 * 0.10 * 20 / 0.02 * 1800
Q = 2.52 * 800 / 36
Q ≈ 90000 J
Therefore, the quantity of heat conducted in 30 min through the iron plate is approximately 90000 J.
The correct answer is (a) 91000000 J.
(Q2) To calculate the thermal conductivity of aluminum, we can use the formula:
k = (Q * d) / (A * ΔT * t)
Where:
k = Thermal conductivity
Q = Quantity of heat conducted
d = Diameter
A = Cross-sectional area
ΔT = Temperature difference
t = Time
Given information:
Q = 164.9 J/s
d = 2.5 cm = 0.025 m
A = π * (d/2)^2 = π * (0.025/2)^2
ΔT = 500°C - 20°C = 480°C
t = 1 s (as Q is given per second)
Substituting the values into the formula:
k = (164.9 * 0.025) / (π * (0.025/2)^2 * 480 * 1)
k ≈ 311.59 W/(m*C)
Therefore, the thermal conductivity of aluminum is approximately 311.6 W/(m*C).
The correct answer is (a) 311.6 W/(m*C).
To determine the quantity of heat conducted in Q1 and the thermal conductivity in Q2, we can use the formula:
Q = (k * A * ΔT * t) / d
where:
Q is the quantity of heat conducted
k is the coefficient of thermal conductivity
A is the area of the surface through which heat is conducted
ΔT is the temperature difference between the two sides
t is the time
d is the thickness of the material
Let's solve these step-by-step:
(Q1) Determine the quantity of heat conducted in 30 min through an iron plate 2.0 cm thick and 0.10 m^2 in area if the temperature of the two sides are 0°C and 20°C. The coefficient of thermal conductivity is 50.4 J/(s*m*C).
Using the formula:
Q = (k * A * ΔT * t) / d
Substituting the given values:
Q = (50.4 J/(s*m*C) * 0.10 m^2 * (20°C - 0°C) * 1800 s) / 2.0 cm
Note: Convert the thickness from cm to m by dividing it by 100.
Q = (50.4 * 0.10 * 20 * 1800) / (2.0/100)
Simplifying the equation gives us:
Q = 50.4 * 0.10 * 20 * 1800 * (100/2.0)
Solving this equation:
Q = 90,000,000 J
Therefore, the quantity of heat conducted is 90,000,000 J. So, the correct answer is (a) 91,000,000 J.
(Q2) One end of a 30 cm long aluminum rod is exposed to a temperature of 500°C, while the other end is maintained at 20°C. The rod has a diameter of 2.5 cm. If heat is conducted through the rod at the rate of 164.9 J/s, calculate the thermal conductivity of aluminum.
Using the formula:
Q = (k * A * ΔT * t) / d
Rearranging the formula to solve for k:
k = (Q * d) / (A * ΔT * t)
Substituting the given values:
k = (164.9 J/s * 30 cm) / (π * (2.5 cm/2)^2 * (500°C - 20°C) * 1 s)
Note: Convert the diameter and length from cm to m by dividing them by 100.
k = (164.9 * 30) / (π * (2.5/2)^2 * (500 - 20))
Simplifying the equation gives us:
k = (164.9 * 30) / (π * 1.25^2 * 480)
Solving this equation:
k ≈ 311.58 W/(m*C)
Therefore, the thermal conductivity of aluminum is approximately 311.58 W/(m*C). So, the correct answer is (a) 311.6 W/(m*C).