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A meat ball with a radius of 25.4 mm at a temperature of 700 K, is suddenly plunged into a medium whose temperature is held at 395 K. Assume a convective heat transfer coefficient of 11.5 W m-2 K-1 and take the average physical properties as: K = 44 W m-1 K-1, ρ = 7850 kg m-3 and
cp = 0.4606 kJ kg-1K-1. The temperature (K) of the meat ball after one hour is _____
    Correct answer is between '472,475'. Can you explain this answer?
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    A meat ball with a radius of 25.4 mm at a temperature of 700 K, is sud...
    Ρ = 7800 kg m-3, and cp = 460 J kg-1 K-1. Calculate the time required for the center of the meatball to reach a temperature of 500 K.

    To solve this problem, we will use the transient conduction equation:

    ρcpV∂T/∂t = kA(∂2T/∂x2) + hA(Tm - T)

    where ρ is the density, cp is the specific heat capacity, V is the volume, T is the temperature, t is time, k is the thermal conductivity, A is the surface area, h is the convective heat transfer coefficient, Tm is the temperature of the medium, and x is the distance.

    First, we need to calculate the volume and surface area of the meatball.

    Volume (V) = (4/3)πr^3 = (4/3)π(25.4 mm)^3 = 4.24 x 10^-5 m^3

    Surface Area (A) = 4πr^2 = 4π(25.4 mm)^2 = 0.081 m^2

    Next, we can rearrange the equation to solve for ∂T/∂t:

    (ρcpV∂T/∂t) = kA(∂2T/∂x2) + hA(Tm - T)

    ∂T/∂t = (kA/(ρcpV))(∂2T/∂x2) + (hA/(ρcpV))(Tm - T)

    Now, we can substitute the values into the equation:

    ∂T/∂t = (44 W m^-1 K^-1 * 0.081 m^2) / (7800 kg m^-3 * 460 J kg^-1 K^-1 * 4.24 x 10^-5 m^3) * (∂2T/∂x2) + (11.5 W m^-2 K^-1 * 0.081 m^2) / (7800 kg m^-3 * 460 J kg^-1 K^-1 * 4.24 x 10^-5 m^3) * (395 K - T)

    Simplifying this equation, we get:

    ∂T/∂t = 1.037 x 10^6 (∂2T/∂x2) + 1.4167 x 10^6 (395 - T)

    Now, we can solve this equation using numerical methods or finite difference methods. However, in this case, we can approximate the equation by assuming that the temperature distribution is one-dimensional and neglecting the second derivative (∂2T/∂x2) term.

    ∂T/∂t = 1.4167 x 10^6 (395 - T)

    Separating variables and integrating, we get:

    ∫d(T) / (395 - T) = ∫1.4167 x 10^6 dt

    -ln|395 - T| = 1.4167 x 10^6 t + C

    Applying the initial condition T(0) = 700 K, we get:

    -ln|395 - 700| = 1.4167 x 10^6 (0) +
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    A meat ball with a radius of 25.4 mm at a temperature of 700 K, is suddenly plunged into a medium whose temperature is held at 395 K. Assume a convective heat transfer coefficient of 11.5 W m-2 K-1 and take the average physical properties as: K = 44 W m-1 K-1, ρ = 7850 kg m-3 andcp = 0.4606 kJ kg-1K-1. The temperature (K) of the meat ball after one hour is _____Correct answer is between '472,475'. Can you explain this answer?
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    A meat ball with a radius of 25.4 mm at a temperature of 700 K, is suddenly plunged into a medium whose temperature is held at 395 K. Assume a convective heat transfer coefficient of 11.5 W m-2 K-1 and take the average physical properties as: K = 44 W m-1 K-1, ρ = 7850 kg m-3 andcp = 0.4606 kJ kg-1K-1. The temperature (K) of the meat ball after one hour is _____Correct answer is between '472,475'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A meat ball with a radius of 25.4 mm at a temperature of 700 K, is suddenly plunged into a medium whose temperature is held at 395 K. Assume a convective heat transfer coefficient of 11.5 W m-2 K-1 and take the average physical properties as: K = 44 W m-1 K-1, ρ = 7850 kg m-3 andcp = 0.4606 kJ kg-1K-1. The temperature (K) of the meat ball after one hour is _____Correct answer is between '472,475'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A meat ball with a radius of 25.4 mm at a temperature of 700 K, is suddenly plunged into a medium whose temperature is held at 395 K. Assume a convective heat transfer coefficient of 11.5 W m-2 K-1 and take the average physical properties as: K = 44 W m-1 K-1, ρ = 7850 kg m-3 andcp = 0.4606 kJ kg-1K-1. The temperature (K) of the meat ball after one hour is _____Correct answer is between '472,475'. Can you explain this answer?.
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