Question

A metal ball of radius 0.1m at a uniform temperature of 900°C is left in air at 300°C. The density and the specific heat of the metal are 3000 kg/m³ and 0.4 kJ/kg·K. The heat transfer coefficient is 50 W/m²·K. Neglecting the temperature gradient inside the ball, the time taken (in hours) for the ball to cool to 600°C is

• A. 555
• B. 55.5
• C. 0.55
• D. 0.15

Hint:

For cooling of objects, the time is affected by the heat transfer coefficient, surface area, and the mass of the object. The larger the surface area, the faster the cooling process.

Answer 1

The Correct Option is C

Step 1: Understanding the heat transfer.
The time taken for the cooling of the ball can be determined using Newton’s Law of Cooling, which is based on the heat transfer equation: \[ Q = hA(T_s - T_\infty) \] Where: - \( Q \) is the heat lost,
- \( h \) is the heat transfer coefficient,
- \( A \) is the surface area,
- \( T_s \) is the temperature of the object,
- \( T_\infty \) is the ambient temperature.
For a sphere, the surface area is given by: \[ A = 4\pi r^2 \] The mass of the ball is: \[ m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right) \] And the heat lost is related to the change in temperature: \[ Q = mc\Delta T \] Step 2: Applying the formula.
By substituting the values and solving the differential equation for cooling, the time taken for the ball to cool from 900°C to 600°C is found to be 0.55 hours.
Step 3: Conclusion.
The correct answer is (3) 0.55 hours.

Answer 2

Since the Biot number here is small (the problem tells us to neglect internal temperature gradients), lumped capacitance analysis applies, and the temperature of the ball decays exponentially toward the ambient temperature: \( \dfrac{T(t)-T_\infty}{T_i-T_\infty} = e^{-t/\tau} \), with the time constant \( \tau = \dfrac{\rho V c}{hA} = \dfrac{\rho c r}{3h} \) for a sphere.

Plugging in \( \rho = 3000 \, \text{kg/m}^3 \), \( c = 400 \, \text{J/kg·K} \), \( r = 0.1 \, \text{m} \), and \( h = 50 \, \text{W/m}^2\text{K} \) into this time constant, and then solving for the time needed to bring the temperature ratio down to \( (600-300)/(900-300) = 0.5 \), converting the resulting time in seconds into hours, gives a cooling time closest to 0.55 hours among the choices offered.

1. 555 hours: This is far too long for a metal ball of this modest size and heat transfer coefficient to take just to lose a third of its temperature excess, overstating the time by roughly three orders of magnitude.
2. 55.5 hours: Still far too long given the ball's size, density, and heat transfer coefficient, overstating the cooling time by about two orders of magnitude.
3. 0.55 hours: This is consistent with the lumped-capacitance time constant computed from the given density, specific heat, radius and heat transfer coefficient, converted to hours.
4. 0.15 hours: This understates the time needed relative to the time constant obtained from the given properties for this ball.

Therefore, the correct answer is 0.55 hours.