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Calculate the amount of heat transferred through a steel plate given its dimensions, temperatures, and thermal conductivity.

Physics

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Answer

1,049,760,000 J\text{1,049,760,000 J}

Step-by-step solution

4 steps

To calculate the amount of heat flow, we use the formula for heat conduction:

Q=k⋅A⋅ΔT⋅tLQ = \frac{k \cdot A \cdot \Delta T \cdot t}{L}

where:
QQ = amount of heat transferred (Joules)
kk = thermal conductivity (K-value)
AA = cross-sectional area
ΔT\Delta T = temperature difference
tt = time
LL = thickness

  1. Identify the given values and convert them to SI units.

    • Thickness, L=10 mm=0.01 mL = 10 \text{ mm} = 0.01 \text{ m}
    • Width, W=0.6 mW = 0.6 \text{ m}
    • Length, H=1 mH = 1 \text{ m}
    • Time, t=30 minutes=30×60 s=1800 st = 30 \text{ minutes} = 30 \times 60 \text{ s} = 1800 \text{ s}
    • Temperature on one side, T1=70∘CT_1 = 70^\circ\text{C}
    • Temperature on the other side, T2=30∘CT_2 = 30^\circ\text{C}
    • K-value of steel, k=243 W/(m⋅K)k = 243 \text{ W/(m}\cdot\text{K)}
  2. Calculate the cross-sectional area (AA).

    The heat flows through the area perpendicular to the thickness.

    A=W×HA = W \times H A=0.6 m×1 mA = 0.6 \text{ m} \times 1 \text{ m} A=0.6 m2A = 0.6 \text{ m}^2
  3. Calculate the temperature difference (ΔT\Delta T).

    ΔT=T1−T2\Delta T = T_1 - T_2 ΔT=70∘C−30∘C\Delta T = 70^\circ\text{C} - 30^\circ\text{C} ΔT=40∘C\Delta T = 40^\circ\text{C}

    Since a change of 1∘C1^\circ\text{C} is equal to a change of 1 K1\text{ K}, ΔT=40 K\Delta T = 40\text{ K}.

  4. Substitute the values into the heat transfer formula.

    Q=243 W/(m⋅K)×0.6 m2×40 K×1800 s0.01 mQ = \frac{243 \text{ W/(m}\cdot\text{K)} \times 0.6 \text{ m}^2 \times 40 \text{ K} \times 1800 \text{ s}}{0.01 \text{ m}} Q=10497600 J0.01Q = \frac{10497600 \text{ J}}{0.01} Q=1049760000 JQ = 1049760000 \text{ J}

    The amount of heat that will flow through the sheet of steel is 1,049,760,000 J\boxed{\text{1,049,760,000 J}}.

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