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.