Question

A slab having a thickness of 4 cm and measuring 25 cm on a side has a 40^0 C temperature difference between its faces. How much heat flows through its per hour? K=0.0025 cal/cm.s.^0 C

Answer 1

Answer on Question #73564, Physics / Molecular Physics | Thermodynamics

Question. A slab having a thickness of $4\, \text{cm}$ and measuring $25\, \text{cm}$ on a side has a $40\,{}^{\circ}\text{C}$ temperature difference between its faces. How much heat flows through its per hour? $k = 0.0025 \frac{\text{cal}}{(\text{cm} \cdot \text{s}^{-1}\text{C})}$.

Given. $\Delta l = 4\, \text{cm}; S = 25 \times 25\, \text{cm}; \Delta T = 40\,{}^{\circ}\text{C}; k = 0.0025 \frac{\text{cal}}{(\text{cm} \cdot \text{s}^{-1}\text{C})}; t = 3600\, \text{s}$.

Find. $Q$

Solution.

According to the equation of thermal conductivity

$$Q = k \cdot \frac{\Delta T}{\Delta l} \cdot S \cdot t$$

we have

$$Q = k \cdot \frac{\Delta T}{\Delta l} \cdot S \cdot t = 0.0025 \cdot \frac{40}{4} \cdot (25 \times 25) \cdot 3600 = 56.25\, \text{kcal}.$$

Answer. $Q = 56.25\, \text{kcal}$.

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