288KJ is conducted across two opposite faces of a 3m cube of temperature gradient 90ºCm\(^{-1}\) in 7200s. Calculate the thermal conductivity.
4.9 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)
5.0 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)
5.2 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)
5.1 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)
Explanation
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My school is actually correct
Q=288Kj
T.G=90°C/m
t=7200sec
K=? d=3 Area=3²=9m²
converting 288Kj=288000j
T.G=∆@/d
90=∆@/3
∆@=270°C. ( this is temperature difference, let's assume 2 different temperature 30°C and 50°C the difference is 50-30=20°C also if you're to convert to K that's 50+273= 323, 30+273=303. to temperature difference will be 323-303=20K so even if you convert or not the interval remains constant) so from the solution ∆@=270°C=270K
Q/t= KA∆@/d
K=Qd/tA∆@
K=288000×3/7200×9×270
K=0.0493= 4.93×10^-2 W/m/K
We are finding thermal conductivity (k).
The formula you need is:
Q = (k A ΔT t) / L
Now we rearrange for k:
k = (Q × L) / (A × ΔT × t)
Now let’s list what we are given:
Heat energy, Q = 288 kJ = 288,000 J
Time, t = 7200 s
Temperature gradient = 90 °C/m (this is ΔT/L, but we’ll handle it properly)
Cube side = 3 m
Step 1: Find area (A)
Heat is flowing across two opposite faces of a cube.
So area of one face:
A = 3 × 3 = 9 m²
Step 2: Find thickness (L)
Heat travels through the cube from one face to the opposite face:
L = 3 m
Step 3: Temperature difference (ΔT)
They gave temperature gradient:
Gradient = 90 °C per m
So:
ΔT = gradient × distance
ΔT = 90 × 3
ΔT = 270 °C
Step 4: Substitute into formula
k = (Q × L) / (A × ΔT × t)
k = (288,000 × 3) / (9 × 270 × 7200)
Step 5: Calculate numerator
288,000 × 3 = 864,000
Step 6: Calculate denominator
9 × 270 = 2430
2430 × 7200 = 17,496,000
Step 7: Final division
k = 864,000 / 17,496,000
k ≈ 0.0494 W/mK
Final answer:
k ≈ 4.94 × 10⁻² W/mK
