All categories

3 years ago

What is the heat-loss rate through the slab if the ground temperature is 5 ∘C while the interior of the house is 25 ∘C?

Physics

1 answer:

Gala2k [10]3 years ago

3 0

Complete question :

A 12 m x 15 m house is built on a 12-cm-thick concrete slab.

What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C

Answer:

3kW

Explanation:

Given the following :

Dimension of house :

Length = 12m

Width = 15m

Thickness of concrete slab (t) = 12cm

t in metres :

100cm = 1m

12cm = (12/100)m

= 0.12m

Ground temperature (Tg) = 5°C

Interior temperature = (Th) = 25°C

Thermal conductivity of concrete (K) is approximately 1 Wm/k

Using the relation:

Q = KA * [ (Th - Tg) / d]

A = Length * width = (12 *15) = 180

Q = (1 * 180) * [(25°C - 5°C) / 0.12]

Q = 180 * (20/0.12)

Q = 180 * 16.6666

Q = 3,000W = 3kW

You might be interested in

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

7 0

3 years ago

If p(a)=0.07692, p(b)=0.25, and probability of a and b. =0.01923, what is probability of a or b. to four decimal places? select

Triss [41]

Answer:

p(a) * p(b) = .01923

p(b) = .01923 / .07692 = .2500

5 0

2 years ago

If a object is at rest, does it mean there are no forces acting on it

Ganezh [65]

Yes that is correct :)

4 0

2 years ago

Read 2 more answers

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago

Please!! List three things that can affect solubility.

Dennis_Churaev [7]

Nature of the solute and solvent, Temperature, and Pressure all affect solubility.

7 0

3 years ago