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3 years ago

What would the final temperature be if 8.94 X 10 3 joules of heat were added to 454 grams of copper at 23.0 o C?

Physics

1 answer:

ozzi3 years ago

8 0

Answer: $74.1^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $8.94\times 10^3$ Joules

m= mass of copper = 454 g

c = specific heat capacity = $0.385J/g^0C$

Initial temperature of the copper = $T_i$ = 23.0°C

Final temperature of the water = $T_f$ = ?

Change in temperature , $\Delta T=T_f-T_i$

Putting in the values, we get:

$8.94\times 10^3=454\times 0.385\times (T_f-23.0)^0C$

$T_f=74.1^0C$

The final temperature of copper will be $74.1^0C$

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The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

3 years ago

4. As Juan is going to take a shower, the soap falls out of the soap dish on to the

LiRa [457]

The coefficient of friction between the soap and the floor is 0.081

If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.

We know that frictional force F = μN where μ = coefficient of friction between soap and floor.

So, μ = F/N

Since F = 40 N and N = W = 493 N,

μ = F/N

μ = 40 N/493 N

μ = 0.081

So, the coefficient of friction between the soap and the floor is 0.081

Learn more about coefficient of friction here:

brainly.com/question/13923375

5 0

3 years ago

Eric has a mass of 70 . He is standing on a scale in an elevator that is accelerating downward at 1.7 . What is the approximate

pychu [463]

Answer:

Explanation:

The value the scale shows is the reaction force to the normal force (they are equal by Newton's 3rd Law) that the scale exerts on Eric.

The forces on Eric are his weight (downward) and this normal force (upward), so we can write the net force over him as (also using Newton's 2nd Law):

$F=W-N=ma$

which means

$N=W-ma=mg-ma=m(g-a)$

and for our values this is:

$N=mg-ma=(70kg)(9.8m/s^2-1.7m/s^2)=567N$

5 0

3 years ago

Change of 5000 c flows through a circuit in one hour. what is the current intensity?

AleksandrR [38]

Answer:

I = 1.38 A

Explanation:

Given that,

Charge, q = 5000 C

Time, t = 1 hour = 3600 s

We need to find the current intensity. The current intensity is equal to the electric charge per unit time. It can be given by :

$I=\dfrac{q}{t}$

Substitute all the values in the above formula

$I=\dfrac{5000\ C}{3600\ s}\\\\=1.38\ A$

So, the current intensity is 1.38 A.

8 0

3 years ago

If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m

Xelga [282]

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is $\frac{135.9}{A}N$

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

Where Force = $\frac{Pressure}{Area}$ we have

Force = $F_{120}=\frac{135.9}{A}N$ .

4 0

3 years ago