Question

An acrylic testing box with dimensions of 22.7 cm x 22.7 cm x 22.7 cm is covered with an unknown insulation material 1.6 cm thick. A 23 W bulb is used to heat the box. You may assume that there is negligible heat loss through the sides of box. If the bulb raises the inside temperature of the box by 8.1 ºC, determine how much heat (Q) was gained from the box measured in joules (J). Round your answer to the nearest hundredth (0.00).

ρ: 1.20 kg/m3 (density of air, Greek letter rho)

Cp: 1000. J/kg °C (specific heat capacity of air)

Answer 1

To determine the amount of heat gained from the box, we can use the formula:

Q = mcΔT

Where:
Q = heat gained (in joules)
m = mass of air inside the box (in kg)
c = specific heat capacity of air (in J/kg·ºC)
ΔT = change in temperature (in ºC)

First, let's find the mass of air inside the box. The volume of the box is given by:

V = L × W × H

V = (22.7 cm) × (22.7 cm) × (22.7 cm)
V = (22.7 cm)^3

Now, let's convert the volume to m^3:

V = (22.7 cm)^3 × (1 m/100 cm)^3
V = (0.227 m)^3

The density of air is given as 1.20 kg/m^3. So, the mass of air inside the box is:

m = ρV
m = (1.20 kg/m^3) × (0.227 m)^3

Now, let's find the change in temperature:

ΔT = 8.1 ºC

The specific heat capacity of air is given as 1000 J/kg·ºC.

Finally, we can calculate the heat gained:

Q = mcΔT

Remember to round your answer to the nearest hundredth.

Answer 2

Okay, let's go through this step-by-step:

Given:

Acrylic box dimensions: 22.7 cm x 22.7 cm x 22.7 cm

Insulation thickness: 1.6 cm

Bulb wattage: 23 W

Temperature increase: 8.1 °C

Density of air (ρ): 1.20 kg/m3

Specific heat capacity of air (Cp): 1000 J/kg °C

To find the heat gained Q:

Find the volume of the acrylic box:

Inner volume = (22.7 - 1.6) x (22.7 - 1.6) x (22.7 - 1.6) cm^3 = 9735 cm^3

Convert to m^3: 9735 cm^3 x (1 m / 100 cm)^3 = 0.009735 m^3

Use density to find inner air mass:

m = ρV = 1.20 kg/m3 x 0.009735 m^3 = 0.01168 kg

Use specific heat capacity and temperature change to find heat gained:

Q = mcΔT

Q = (0.01168 kg)(1000 J/kg°C)(8.1 °C)

Q = 94.5 J

Rounded to the nearest hundredth, the heat gained is 94.50 J.

Step-by-step explanation:

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