Question

Problem 3 - An acrylic testing box is covered on the top with an unknown 15 mm insulation material with dimensions of 25 cm by 25 cm. A 25.0W bulb is used to heat the box and it maintains the inside temperature at 10.0ºC higher than the outside temperature.

a. Determine the thermal conductivity (k) for the insulating material.
b. Find the energy transferred (Q) through the material after a half hour.
c. Find the R-value of the insulation.


I don't understand how to do this

Answer 1

a. 0.6 W/mK b. 45 kJ c. 0.025 m²K/W

Step-by-step explanation:

a. Determine the thermal conductivity (k) for the insulating material.

The rate at which heat is transferred across the insulating material P = kAΔT/L where K = thermal conductivity, A = area of insulating material = 25 cm × 25 cm = 0.25 m × 0.25 m = 0.0625 m², ΔT = temperature difference between inside and outside = 10.0 °C = 10.0 K and L = thickness of insulating material = 15 mm = 0.015 m

So, making k subject of the formula, we have

k = PL/AΔT

Now, P = 25.0 W = heat delivered to box = heat delivered to insulating material per second.

k = 25.0 W × 0.015 m/(0.0625 m² × 10.0 K)

k = 0.375 Wm/(0.625 m²K)

k = 0.6 W/mK

b. Find the energy transferred (Q) through the material after a half hour.

Since quantity of energy transferred Q = Pt where P = heat delivered to box = heat delivered to insulating material per second = 25.0 W and t = time the heat is delivered = 1/2 h = 0.5 h = 0.5 × 60 min = 30 min × 60 s/min = 1800 s

So, Q = 25.0 W × 1800 s

Q = 45000 J

Q = 45 kJ

c. Find the R-value of the insulation.

R - value, R = L/k = 0.015 m/0.6 W/mK = 0.025 m²K/W