Determine the time required for the center temperature of a 3 cube to reach 808C. The cube has a volume of 125 cm . The thermal conductivity of the material is 0.4 W/(m 8C); density is 950 kg/m3; and specific heat is 3.4 kJ/(kg K). The initial temperature is 208C. The surrounding temperature is 908C. The cube is immersed in a fluid that results in a negligible surface resistance to heat transfer.

The correct question with clearer values is;

Determine the time required for center temperature of cube to reach 80°C. the cube has volume of 125 cm³. Thermal conductivity of material is 0.4 W/(m C); density is 950 kg/m3, and specific heat is 3.4 kJ/(kg K). the initial temperature is 20°C. The surrounding temperature is 90°C. The cube is immersed in a fluid that results in negligible surface resistance to heat transfer.

Time required = 1 hour

Explanation:

An approximate answer can be obtained by assuming linear heat conduction from surface to core. The total heat transferred to the cube is given as;

Q=mc•ΔT.

We need the mass of the cube.

We know that mass /volume = density.

Thus, mass = volume x denisty

We are given;

specific heat;c = 3.4 kJ/kg.K = 3400 kJ/kg.K

Volume = 950 kg/m³

Density = 125 cm³ = 125 x 10^(-6) m³

So,

Mass = 950 x (125 x 10^(-6)) = 0.1188 kg

Now, when the core has reached 80°C, let's assume that the average cube temperature is 85°C so that ΔT = 85 - 20 = 65 °C.

Thus;

Q = 0.1188 x 3400 x 65 = 26250 J

Now, to find the time for the core to reach 80°C from the heat equation,we will use the 1-dimensional heat conduction equation;

Q/t = k•A•ΔT'/L

Let's make t the subject

t= QL/(k•A•ΔT')

We know that the volume of the cube is 125 cm³

Thus, its side length is ∛125 = 5 cm

surface area of a cube is given by the formula; A = 6L²

Thus,A = 6 x 5 x 5 = 150 cm² =0.015 m².

Now, Let's take for L the shortest surface-core distance, 2.5 cm=0.025m and let's assume the average surface temperature during the conduction process is (90+20)/2=55°C, while the average core temperature is 25°C because it increases exponentially.

Thus, ΔT'= 55 - 25 = 30°C,

Plugging in the relevant values into;

t= QL/(k•A•ΔT'), we have;

t= (26250 x 0.025)/(0.4x0.015x30)

t = 3645.83 seconds

Converting this to hours, we know that 3600 seconds make 1 hour.

Thus, t = 3645.83/3600 = 1.0127 hours.

This is approximately 1 hour

3646 s ≈ 1 hour

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