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.
Answer:
Time required = 1 hour
Step-by-step 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