A brass rod (thermal conductivity 109 J/s m K) has an area of cross section 0.04 m 2and length 20 cm. If the two end of the rod are maintained at a temperature difference of 200°C, the rate of heat flow through the rod is ________.
The question asks us to calculate the rate of heat flow through a brass rod given its dimensions, thermal conductivity, and the temperature difference across its ends. This is a classic application of Fourier's Law of Heat Conduction.
Fourier's Law of Heat Conduction:
The rate of heat flow (\(\frac{Q}{t}\)) through a material is directly proportional to the area of cross-section (\(A\)), the temperature difference (\(\Delta T\)), and the thermal conductivity (\(k\)) of the material, and inversely proportional to the length (\(L\)) of the material. The formula is given by:
\(\frac{Q}{t} = k A \frac{\Delta T}{L}\)
where:
Let's list the values provided in the question:
Before plugging the values into the formula, we need to ensure all units are consistent. The length is given in centimeters, so we convert it to meters:
\(L = 20 \, \text{cm} = 20 \times 10^{-2} \, \text{m} = 0.20 \, \text{m}\)
The temperature difference is given in °C. For a temperature difference, the value in Kelvin is the same as in Celsius. So, \(\Delta T = 200 \, \text{K}\).
Now, substitute the values into the formula for the rate of heat flow:
\(\frac{Q}{t} = k A \frac{\Delta T}{L}\)
\(\frac{Q}{t} = (109 \, \text{J/s m K}) \times (0.04 \, \text{m}^2) \times \frac{200 \, \text{K}}{0.20 \, \text{m}}\)
Let's calculate the numerical value:
\(\frac{Q}{t} = 109 \times 0.04 \times \frac{200}{0.2}\)
\(\frac{Q}{t} = 109 \times 0.04 \times 1000\)
\(\frac{Q}{t} = 109 \times 40\)
\(\frac{Q}{t} = 4360 \, \text{J/s}\)
The options are given in kilojoules per second (kJ/s). We convert the result from J/s to kJ/s:
\(1 \, \text{kJ} = 1000 \, \text{J}\)
So,
\(4360 \, \text{J/s} = \frac{4360}{1000} \, \text{kJ/s} = 4.360 \, \text{kJ/s}\)
The rate of heat flow through the brass rod is 4.36 kJ/s.
Let's compare our calculated rate of heat flow with the given options:
Our calculated value of 4.36 kJ/s matches Option 3.
| Concept | Description | Formula |
|---|---|---|
| Heat Conduction | Transfer of heat through direct contact, occurring in solids, liquids, and gases (most significant in solids). | N/A |
| Thermal Conductivity (\(k\)) | A material's ability to conduct heat. Higher \(k\) means better conductor. | Units: J/s m K or W/m K |
| Rate of Heat Flow (\(\frac{Q}{t}\)) | The amount of heat energy transferred per unit time. | \(\frac{Q}{t} = k A \frac{\Delta T}{L}\) (for steady-state conduction) |
| Temperature Gradient (\(\frac{\Delta T}{L}\)) | The change in temperature per unit length along the direction of heat flow. | Units: K/m or °C/m |
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