How long does it take for a tea cup with a curved surface area of 35 cm², open to air on the top, to cool from 90°C to 30°C in a room at 20°C? Consider real-world factors like convection, radiation, and mild air circulation. Assume the heat transfer coefficient for convection is 15 W/m²K, the emissivity of the tea is 0.95, and the tea has a mass of 200 g with a specific heat capacity of 4186 J/kg·K. The top of the tea cup is circular with a radius of 4 cm.
The total surface area consists of the side surface area and the open top area:
Given the curved side surface area \( A_{\text{side}} = 35 \, \text{cm}^2 = 0.0035 \, \text{m}^2 \), and the radius of the top of the cup \( r = 4 \, \text{cm} = 0.04 \, \text{m} \), the area of the top is:
\[ A_{\text{top}} = \pi r^2 = \pi \times (0.04)^2 = 0.00503 \, \text{m}^2 \]
So, the total surface area exposed to cooling is:
\[ A_{\text{total}} = A_{\text{side}} + A_{\text{top}} = 0.0035 + 0.00503 = 0.00853 \, \text{m}^2 \]
We calculate the convective heat loss using Newton's law of cooling:
\[ q_{\text{conv}} = h A_{\text{total}} (T_{\text{tea}} - T_{\text{room}}) \]
Substituting the values:
\[ q_{\text{conv}} = 15 \times 0.00853 \times (T_{\text{tea}} - 20) = 0.12795 \times (T_{\text{tea}} - 20) \, \text{W} \]
The radiative heat loss is calculated using Stefan-Boltzmann's law:
\[ q_{\text{rad}} = \epsilon \sigma A_{\text{total}} \left( T_{\text{tea}}^4 - T_{\text{room}}^4 \right) \]
Where:
Substitute the values into the formula:
\[ q_{\text{rad}} = 0.95 \times 5.67 \times 10^{-8} \times 0.00853 \times \left( 363.15^4 - 293.15^4 \right) \]
First, calculate the fourth powers of the temperatures:
\[ 363.15^4 = 1.736 \times 10^8, \quad 293.15^4 = 7.398 \times 10^7 \]
Thus, the radiative heat loss becomes:
\[ q_{\text{rad}} = 0.95 \times 5.67 \times 10^{-8} \times 0.00853 \times (1.736 \times 10^8 - 7.398 \times 10^7) \]
Finally, simplifying the expression:
\[ q_{\text{rad}} \approx 0.0458 \, \text{W} \]
The total heat loss is the sum of convective and radiative heat losses:
\[ q_{\text{total}} = q_{\text{conv}} + q_{\text{rad}} = 0.12795 \times (T_{\text{tea}} - 20) + 0.0458 \]
The energy contained in the tea initially is given by:
\[ Q = mc(T_{\text{tea}} - T_{\text{room}}) \]
Substitute the known values:
\[ Q = 0.2 \times 4186 \times (90 - 20) = 58604 \, \text{J} \]
The rate of temperature change is given by the differential equation:
\[ \frac{dT}{dt} = -\frac{q_{\text{total}}}{mc} \]
Substitute the expression for total heat loss:
\[ \frac{dT}{dt} = -\frac{1}{mc} \left[ 0.12795 \times (T_{\text{tea}} - 20) + 0.0458 \right] \]
With \( m = 0.2 \, \text{kg} \) and \( c = 4186 \, \text{J/kg} \cdot \text{K} \), the equation becomes:
\[ \frac{dT}{dt} = -\frac{1}{0.2 \times 4186} \left[ 0.12795 \times (T_{\text{tea}} - 20) + 0.0458 \right] \]
Now we solve this equation numerically to find the time required for the tea to cool from 90°C to 30°C.
Using numerical integration methods (like Euler's method or Runge-Kutta), we find that the tea will take approximately 3 to 4 hours to cool from 90°C to 30°C in a room at 20°C, accounting for convection, radiation, and air circulation.