The rate of heat transfer Q across a building wall varies directly as the surface area A o… — JAMB UTME

The rate of heat transfer $Q$ across a building wall varies directly as the surface area $A$ of the wall and the temperature difference $\Delta T$ between the interior and exterior, and inversely as the wall thickness $d$ . When the surface area is $4\text{ m}^2$ , the temperature difference is $15^\circ\text{C}$ , and the thickness is $0.05\text{ m}$ , the heat transfer rate is $1200\text{ W}$ . What is the heat transfer rate in watts when the surface area is $6\text{ m}^2$ , the temperature difference is $20^\circ\text{C}$ , and the thickness is $0.08\text{ m}$ ?

Cevap: 1500 W

Cevap

The heat transfer rate is

1500\text{ W}

Establishing the variation constant

k = 1

using the initial given values and substituting the new parameters yields

Q = \frac{1 \times 6 \times 20}{0.08} = 1500\text{ W}

Adım Adım Çözüm

Formulate the variation equation

Q = \frac{k A \Delta T}{d}

Direct variation means multiplying by

A

and

\Delta T

, while inverse variation means dividing by

d

Calculate the constant of variation

k

k = 1

Substituting

Q=1200

A=4

\Delta T=15

, and

d=0.05

gives

1200 = \frac{60k}{0.05} = 1200k

, so

k = 1

Compute the target heat transfer rate

Q

1500\text{ W}

Substituting

k=1

A=6

\Delta T=20

, and

d=0.08

gives

Q = \frac{1 \times 6 \times 20}{0.08} = \frac{120}{0.08} = 1500\text{ W}

Anahtar Kavram

Joint and Inverse Variation

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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