A metal ball of radius 0.1 m at a uniform temperature of 90°C is left in air at 30°C. The density and the specific heat of the metal are 3000 kg/m³ and 0.4 kJ/kg.K respectively. The heat transfer co-efficient is 50 W/m².K Neglecting the temperature gradients inside the ball, the time taken (in hours) for the ball to cool to 60°C is

ANS:D - 0.15

To find the time taken for the metal ball to cool from 90°C to 60°C, we can use Newton's law of cooling, which states that the rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings. The rate of heat loss can be expressed as: =ℎ(−ambient)dtdQ​=hA(T−Tambient​) Where:

• dtdQ​ is the rate of heat loss (W),
• ℎh is the heat transfer coefficient (W/m²·K),
• A is the surface area of the metal ball (m²),
• T is the temperature of the metal ball (°C), and
• ambientTambient​ is the ambient temperature (°C).

The rate of heat loss is also equal to the change in internal energy of the metal ball: dtdQ​=mcdtdT​ Where:

• m is the mass of the metal ball (kg),
• c is the specific heat of the metal (kJ/kg·K), and
• dtdT​ is the rate of change of temperature (°C/h).

We can set these two expressions equal to each other: ambient)mcdtdT​=hA(T−Tambient​) We can solve this equation for dtdT​ and integrate it to find the time taken for the temperature to change from 90°C to 60°C. Given:

• Radius of the metal ball, =0.1r=0.1 m
• Density of the metal, =3000ρ=3000 kg/m³
• Specific heat of the metal, =0.4c=0.4 kJ/kg·K
• Ambient temperature, ambient=30Tambient​=30°C
• Heat transfer coefficient, ℎ=50h=50 W/m²·K
• Final temperature, =60Tf​=60°C
• Initial temperature, =90Ti​=90°C

First, let's find the mass and surface area of the metal ball: =433V=34​πr3 m=ρV=42A=4πr2 Substitute the given values: =43(0.1)3≈4.188×10−3 m3V=34​π(0.1)3≈4.188×10−3m3 =3000×4.188×10−3≈12.564 kgm=3000×4.188×10−3≈12.564kg =4(0.1)2=4×0.01≈0.1257 m2A=4π(0.1)2=4π×0.01≈0.1257m2 Now, rearrange the equation for dtdT​: =ℎ(−ambient)dtdT​=mchA(T−Tambient​)​ Substitute the given values: =50×0.1257×(90−30)12.564×0.4dtdT​=12.564×0.450×0.1257×(90−30)​ ≈50×0.1257×6012.564×0.4dtdT​≈12.564×0.450×0.1257×60​ ≈50×7.54212.564×0.4dtdT​≈12.564×0.450×7.542​ ≈377.15.0256dtdT​≈5.0256377.1​ ≈75 °C/hdtdT​≈75°C/h Now, we have the rate of change of temperature. To find the time taken for the temperature to change from 90°C to 60°C, we'll integrate dtdT​ with respect to time: =∫075 ∫Ti​Tf​​dT=∫0t​75dt =75Tf​−Ti​=75t 60−90=60−90=75t −30=75−30=75t =−3075t=−7530​ −0.4 hourst=−0.4hours Since time cannot be negative, there might be a mistake in the calculations. Let's check our work and correct any errors.