Please help I have until 9 to turn in.

what is the energy required to melt down a 12,400g Gold bar assuming its temperature goes from 23c to 1,064c

sergejj [24]

1665183.6J

Explanation:

Given parameters:

Mass of Gold = 12400g

Initial temperature = 23°C

Final temperature = 1064°C

Unknown:

Energy required to melt = ?

Solution:

The amount of heat required to cause this temperature change to heat the gold bar is given as;

H = m c Ф

m is the mass of gold bar

c is the specific heat capacity of gold

Ф is the heat change

Specific heat capacity of gold = 0.129J/g°C

H = m c (Ф₂ - Ф₁)

H = 12400 x 0.129 x (1064 - 23) = 1665183.6J

learn more:

Specific heat capacity brainly.com/question/7210400

#learnwithBrainly

5 0

3 years ago

A cyclist has a constant speed of 12 m/s. What is the magnitude of the displacement of the cyclist after 18 seconds?

Katyanochek1 [597]

Answer:

216 m

Explanation:

Assuming a straight line:

Δx = vt

Δx = (12 m/s) (18 s)

Δx = 216 m

7 0

3 years ago

A change in position of an object is its:

finlep [7]

Answer:

it's B I think I did this the other day tbh sorry if it's wrong btw hope this helps

8 0

3 years ago

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 75-kg (165-lb) person o

CaHeK987 [17]

Answer:

(a) Q = 142.67 W

(b) Basal Metabolic Rate = 178.33 W

Explanation:

(a)

We can find the heat radiated by the person by using Stefan-Boltzman's law:

$Q = \sigma A (T^4 - T_{s}^4)\\$

where,

Q = heat radiated per second = ?

σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

A = Surface Area = 2 m²

T = Temperature of Skin = 30° C + 273 = 303 k

Ts = Temperature of room = 18° C +273 = 291 k

Therefore,

$Q = (5.6703\ x\ 10^{-8}\ W/m^2.k^4)(2\ m^2)[(303\ k)^4-(291\ k)^4]$

Q = 142.67 W

(b)

Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,

$Q = (0.8)(Basal\ Metabolic\ Rate)\\Basal\ Metabolic\ Rate = \frac{Q}{0.8}\\\\Basal\ Metabolic\ Rate = \frac{142.67\ W}{0.8}$

Basal Metabolic Rate = 178.33 W

5 0

3 years ago

A wire in a circuit carries a current of 0.9 A. Calculate the quantity of

raketka [301]

Answer:

We conclude that the quantity of the charge that flows through the wire in 50 s will be 45 C.

Explanation:

Given

Current I = 0.9 A

Time t = 50 s

To determine

We need to find the quantity of the charge that flows through the wire in 50 s.

Important Tip:

A current of 1 Ampere = 1 Coulomb of charge flowing in 1 second

Using the formula involving charge and current

$I=\frac{Q}{t}$

where:

$I$ represents the current in amperes (A)

$Q$ represents the charge in coulomb (C)

$t$ represents the time in seconds (s)

now substituting I = 0.9 and t = 50 in the formula

$I=\frac{Q}{t}$

$0.9\:=\:\frac{Q}{50}$

switch sides

$\frac{Q}{50}=0.9$

Multiply both sides by 50

$\frac{50Q}{50}=0.9\cdot \:50$

Simplify

$Q=45$ C

Therefore, we conclude that the quantity of the charge that flows through the wire in 50 s will be 45 C.

6 0

3 years ago