Subtopic: Transient heat transfer in simple geometry
Question Statement:
A small cylindrical sausage of mass $m = 0.050\text{ kg}$, surface area $A = 0.0060\text{ m}^2$, and specific heat capacity $C_p = 3500\text{ J/(kg}\cdot\text{K)}$ is immersed in an agitated boiling water bath maintained at $T_\infty = 100.0^\circ\text{C}$. The convective heat transfer coefficient is $h = 120.0\text{ W/(m}^2\cdot\text{K)}$. Assuming lumped capacitance behavior ($\theta = \frac{T - T_\infty}{T_0 - T_\infty} = e^{-t / \tau}$), the thermal time constant $\tau = \frac{m C_p}{h A}$ in seconds is ________ (round off to one decimal place).
Answer Key:
243.1
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Direct Time Constant Evaluation**
1. Calculate total thermal capacitance:
$C_{\text{th}} = m \times C_p = 0.050\text{ kg} \times 3500\text{ J/(kg}\cdot\text{K)} = 175.0\text{ J/K}$
2. Calculate thermal conductance:
$K_{\text{th}} = h \times A = 120.0\text{ W/(m}^2\cdot\text{K)} \times 0.0060\text{ m}^2 = 0.720\text{ W/K}$
3. Thermal time constant:
$\tau = \frac{C_{\text{th}}}{K_{\text{th}}} = \frac{175.0}{0.720} = 243.055\text{ s} \approx 243.1\text{ s}$
**Method 2: Formula Form**
$\tau = \frac{m C_p}{h A} = \frac{0.050 \times 3500}{120.0 \times 0.0060} = \frac{175}{0.72} = 243.06\text{ seconds}$