Topic: Thermodynamics / Fluid Mechanics
A weather balloon is rising vertically in the atmosphere. Temperature (T) at any point depends on both time (t) and altitude (h). The balloon rises with a vertical velocity of $10~m/s$. For a given time the change in temperature with respect to altitude is $-0.01^{\circ}C/m$. Similarly for a given altitude, the change in temperature with respect to time is $-0.1^{\circ}C/s$. The total change in temperature $(dT/dt)$ experienced by the balloon is:
- A. $-0.2^{\circ}C/s$
- B. $-0.11^{\circ}C/s$
- C. $-0.09^{\circ}C/s$
- D. $-0.1^{\circ}C/s$
Answer: A. $-0.2^{\circ}C/s$
Explanation: The total derivative of temperature with respect to time (material derivative) is $\frac{dT}{dt} = \frac{\partial T}{\partial t} + v \frac{\partial T}{\partial h}$.
Given: Local rate of change $\frac{\partial T}{\partial t} = -0.1^{\circ}C/s$.
Vertical velocity $v = \frac{dh}{dt} = 10~m/s$.
Temperature gradient $\frac{\partial T}{\partial h} = -0.01^{\circ}C/m$.
$\frac{dT}{dt} = -0.1 + (10 \times -0.01) = -0.1 - 0.1 = -0.2^{\circ}C/s$.