Ask question

Ask question

All categories

2 years ago

10

The specific heat of water is 4190 j/(kg*k). suppose you put 1 kg of water (a bit over 4 cups) into a microwave that can deliver

1000 w of power to the water when it is running at full power. roughly how long will you need to run the microwave at full power in order to raise the temperature of the water by 10Â°c?

1 answer:

VladimirAG [237]2 years ago

4 0

Given:
P = 1000 W, power input
c = 4190 J/(kg-K), the specific heat of water
m = 1 kg, mass of water
ΔT = 10 °C = 10 K, temperature rise.

Let t = time required to raise the temperature f the water.
Then
P*t = m*c*ΔT
(1000 J/s)*(t s) = (1 kg)*(4190 J/(kg-K)*(10 K)
1000t = 41900
t = 41.9 s

Answer: 41.9 s or 42 s (approximately)

You might be interested in

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

Rashid [163]

I believe the answer is #4. u can always ask google if u believe that's the wrong answer :)

4 0

2 years ago

Read 2 more answers

Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio

Pachacha [2.7K]

First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N

Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>

5 0

2 years ago

Read 2 more answers

Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre

alex41 [277]

Answer:

The simplified expression is $M = \frac{v^2 r}{G}$

Explanation:

From the question we are told that

$M = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }$

So simplifying we have

$M = \frac{m v^2}{r} * \frac{r^2 }{ mG }$

$M = \frac{v^2 r}{G}$

Thus the simplified formula is $M = \frac{v^2 r}{G}$

3 0

2 years ago

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

How much power does a machine have that does 15204 J of work in 40 seconds?

pentagon [3]

Explanation:

since P=W/t

P=15204/40

P=380.1 Watt

5 0

1 year ago