Problem Statement

A mug of hot tea is cooling to ambient temperature through convective heat transfer with the surrounding air. The cooling process is influenced by both:

• The air in contact with the top surface of the tea
• The mug walls in contact with the surrounding air

1. A room with normal airflow and no additional forced convection.
2. A room with a ceiling fan that provides additional airflow, thus increasing the convective heat transfer rate.

Derivation of Cooling Time

Let’s define the parameters and variables:

• \( T(t) \): Temperature of the tea at time \( t \)
• \( T_{\text{amb}} \): Ambient temperature
• \( T_0 \): Initial temperature of the tea
• \( h \): Convective heat transfer coefficient without fan
• \( h_{\text{fan}} \): Convective heat transfer coefficient with fan
• \( A \): Surface area of tea in contact with air
• \( m \): Mass of tea
• \( c \): Specific heat capacity of tea

Newton’s Law of Cooling

Using Newton's Law of Cooling, the rate of heat loss can be modeled as: \[ \frac{dQ}{dt} = -h A (T(t) - T_{\text{amb}}) \] Where \( Q \) is the thermal energy.

The temperature change over time \( T(t) \) can be expressed as:

\[ m c \frac{dT}{dt} = -h A (T(t) - T_{\text{amb}}) \] Rearranging and solving this differential equation for temperature \( T(t) \) gives: \[ T(t) = T_{\text{amb}} + (T_0 - T_{\text{amb}}) e^{-\frac{h A}{m c} t} \]

Cooling Time with Ceiling Fan

With a ceiling fan providing forced convection, the convective heat transfer coefficient \( h \) increases to \( h_{\text{fan}} \). This affects the rate of cooling as follows: \[ T_{\text{fan}}(t) = T_{\text{amb}} + (T_0 - T_{\text{amb}}) e^{-\frac{h_{\text{fan}} A}{m c} t} \]

Comparison of Cooling Times

The ratio of cooling times can be approximated as:

\[ \frac{t_{\text{fan}}}{t} = \frac{\frac{h A}{m c}}{\frac{h_{\text{fan}} A}{m c}} = \frac{h}{h_{\text{fan}}} \]

Graphical Representation

The graph shows the faster cooling rate with the fan due to increased convective heat transfer.