Question

An 18-cm-long, 16-cm-wide, and 12-cm-high hot iron block $\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( initially at \)20^{\circ} \mathrm{C}$ is placed in an oven for heat treatment. The heat transfer coefficient on the surface of the block is \(100 \mathrm{~W} / \mathrm{m}^{2} . \mathrm{K}\). If the temperature of the block must rise to \(750^{\circ} \mathrm{C}\) in a 25 -min period, the oven must be maintained at (a) \(750^{\circ} \mathrm{C}\) (b) \(830^{\circ} \mathrm{C}\) (c) \(875^{\circ} \mathrm{C}\) (d) \(910^{\circ} \mathrm{C}\) (e) \(1000^{\circ} \mathrm{C}\)

Short answer

Answer choices: A) 618°C B) 829°C C) 975°C D) 1020°C

Step-by-step solution

Calculate the volume and mass of the iron block

First, we need to calculate the volume of the iron block by using the formula for volume of a rectangular block: \(V = L \times W \times H\) Where L = 18 cm, W = 16 cm, and H = 12 cm. Plug the values into the equation and convert the volume from cm³ to m³ by multiplying the result by \(10^{-6}\). Next, find the mass of the iron block by using \(mass = \rho \times V\) Where \(\rho\) is the density of the iron block, which is 7870 kg/m³.

Calculate the total heat transfer required

To calculate the total heat transfer needed to raise the block's temperature to the desired value, use the equation: \(Q = mass \times c_p \times \Delta T\) Where \(Q\) is the total heat transfer, \(c_p\) is the specific heat capacity of iron (447 J/kg·K), and \(\Delta T\) is the change in temperature, which is 750°C - 20°C.

Determine the heat transfer rate

Now, we need to calculate the heat transfer rate, \(q\), needed to raise the temperature of the block within the given time frame. To do this, divide the total heat transfer by the time in seconds: \(q = \frac{Q}{time}\) Where time is 25 minutes, converted to seconds by multiplying by 60.

Calculate the required oven temperature

Finally, we need to calculate the required oven temperature using Newton's Law of Cooling: \(q = hA(\Theta_{oven} - \Theta_{block})\) Where \(q\) is the heat transfer rate, \(h\) is the heat transfer coefficient (100 W/m²·K), \(A\) is the surface area of the block, and \(\Theta_{oven}\) and \(\Theta_{block}\) are the oven and block temperatures in Kelvin, respectively. We assume a uniform temperature across the block, so we can use the final block temperature (750°C). First, find the surface area using the formula \(A = 2(LW + LH + WH)\), where L = 0.18 m, W = 0.16 m, and H = 0.12 m. Next, rearrange the equation to solve for \(\Theta_{oven}\): \(\Theta_{oven} = \frac{q}{hA} + \Theta_{block}\) Plug the values into the equation and make sure to use the Kelvin temperature scale. Find the oven temperature required to heat the block to 750°C within 25 minutes, and choose the appropriate option from the given choices.