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soldi70 [24.7K]

3 years ago

15

The iron nail’s mass is 16 grams and its temperature drops 650 C when dropped into the water. How much heat energy did the iron

nail transfer to the water?

1 answer:

Mice21 [21]3 years ago

5 0

The heat energy transferred by the iron nail is 4680 J

Explanation:

The thermal energy transferred by a substance to another substance is given by the equation

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is its change in temperature

For the iron nail in this problem, we have:

m = 16 g

$C=0.450 J/g^{\circ}C$

$\Delta T = -650^{\circ}C$

So, the amount of heat energy given off by the nail is

$Q=(16)(0.450)(-650)=-4680 J$

where the negative sign indicates that the heat is given off.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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Q.1 Lorna built the circuit diagram below. All the bulbs are identical. (a) Complete the table below by writing on or off for ea

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Answer:

on and off

Explanation:

if there are switches, it can change if the electricity can get to the bulb or not. if it appears that there is no pathway for the electricity to get to the light bulb, it is of, if there is a pathway, its on

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2 years ago

PLEASE HELP!!!!

VashaNatasha [74]

In my opinion I would say all of the above; because people's job depends on where they live or how their environment is. The way they travel is also affected because we can't travel by car through water and can't use a boat in land. Free time as well because people like to travel and or just stay at home depending on the weather which is part of earths features.

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3 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

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3 0

3 years ago

A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin

blsea [12.9K]

Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$ (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$

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Answer:

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Explanation:

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2 years ago

Read 2 more answers