Ask question

Ask question

All categories

3 years ago

11

To make yourself some coffee, you put one cup of water (246 gg ) in a small pot on the stove. Part A What quantity of energy mus

t be transferred thermally to the water to raise its temperature from 20 ∘C∘C to 100 ∘C∘C?

1 answer:

Leya [2.2K]3 years ago

3 0

Answer:

The required heat energy is $\bf{19680~cal}$ .

Explanation:

The heat energy 'Q' required to raise the temperature of water is given by

$Q = M~s~\Delta \theta$

where 'M' is the mass of water, 's' is the specific heat capacity of water and ' $\Delta \theta$ ' is the change of temperature.

Given, $M = 246 gm, \Delta \theta = (100^{0}~C - 20^{0}~C) = 80^{0}~C$ and we know that the specific heat capacity of water is $s = 1~cal~gm^{-1}~^{0}C^{-1}$ .

Substituting the values in the above expression, the required heat energy is

$Q = 246~gm \times 1~cal~gm^{-1}~^{0}C^{-1} \times 80^{0}~C = 19680~Cal$

You might be interested in

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

(c) the maximum speed attained by the object during its motion

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

3 years ago

Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?

Debora [2.8K]

$493 \; \text{W}\cdot \text{m}^{-2}$ .

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

$P$ the total power emitted,
$A$ the surface area of the body,
$\sigma$ the Stefan-Boltzmann Constant, and
$T$ the temperature of the body in degrees Kelvins.

$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$ .

$T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}$ .

$\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}$ .

Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.

5 0

3 years ago

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s

ad-work [718]

Answer:

a

$\theta = 0.0022 rad$

b

$I = 0.000304 I_o$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

The distance of the slit separation is $d = 0.500 \ mm = 5.0 *10^{-4} \ m$

Generally the condition for two slit interference is

$dsin \theta = m \lambda$

Where m is the order which is given from the question as m = 2

=> $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

substituting values

$\theta = 0.0022 rad$

Now on the second question

The distance of separation of the slit is

$d = 0.300 \ mm = 3.0 *10^{-4} \ m$

The intensity at the the angular position in part "a" is mathematically evaluated as

$I = I_o [\frac{sin \beta}{\beta} ]^2$

Where $\beta$ is mathematically evaluated as

$\beta = \frac{\pi * d * sin(\theta )}{\lambda }$

substituting values

$\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }$

$\beta = 0.06581$

So the intensity is

$I = I_o [\frac{sin (0.06581)}{0.06581} ]^2$

$I = 0.000304 I_o$

3 0

3 years ago

What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?

salantis [7]

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the speed over the ground is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

For more information on this visit

brainly.com/question/22271063?referrer=searchResults

8 0

3 years ago

Read 2 more answers

Can someone help me please

TEA [102]

Answer:

D

Explanation:

Because it is impossible for it to show the real depth of the ocean and how deep it is

8 0

3 years ago