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3 years ago

How much heat is absorbed by 60 g of copper (c=385 J/kg K) when it is heated

Physics

1 answer:

MArishka [77]3 years ago

3 0

Answer:

Q = 1386 J

Explanation:

Mass, m = 60 g = 0.06 kg

Specific heat of copper, c=385 J/kg K

It is heated from 20°C to 80°C.

We need to find the heat absorbed. It can be given by the formula as follows :

$Q=mc\Delta T\\\\Q=0.06\times 385 \times (80-20)\\\\Q=1386\ J$

So, 1386 J of heat is absorbed.

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PLEASE HELP!

WINSTONCH [101]

So you would first multiply 400 by 2 which equals 800, then add 30 which is 830.
Then you would subtract 1000-830=170.
The total force of the 6 other players would be 170N.
Hoped this helped ☺️

7 0

3 years ago

Just think about this.

diamong [38]

This ain’t the place, bud. If you have a QUESTION, then you can post it here.

8 0

3 years ago

Read 2 more answers

Calculate the heat energy released when 13.3 g of liquid mercury at 25.00 C is converted to solid mercury at its melting point.C

dmitriy555 [2]

Answer:

$-270.321012\ J$

Explanation:

$C_v$ = Heat capacity of Hg = 28 J/mol

$\Delta T$ = Change in temperature = $(234.32-(273.15+25))$

$\Delta H_{f}$ = Enthalpy of fusion = 2.29 kJ/mol

The number of moles is given by

$n=13.3\times \dfrac{1}{200.59}\\\Rightarrow n=0.0663\ molHg$

Heat is given by

$Q_1=nC_v\Delta T\\\Rightarrow Q_1=0.0663\times 28\times (234.32-(273.15+25))\\\Rightarrow Q_1=-118.494012\ J$

Heat released is given by

$Q_2=-n\Delta H_{f}\\\Rightarrow Q_2=-0.0663\times 2.29\times 10^3\\\Rightarrow Q_2=-151.827\ J$

Total heat is given by

$Q=Q_1+Q_2\\\Rightarrow Q=-118.494012+(-151.827)\\\Rightarrow Q=-270.321012\ J$

The total heat released is $-270.321012\ J$

4 0

3 years ago

A mass M suspended by a spring with force constant k has aperiod T when set into oscillations on Earth. Its period on Mars,whose

olchik [2.2K]

Answer:

C)T

Explanation:

The period of a mass-spring system is:

$T=2\pi\sqrt\frac{m}{k}$

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

6 0

4 years ago

A hot-air balloon is rising straight up with a speed of 1.30 m/s. A ballast bag is released from rest relative to the balloon wh

DiKsa [7]

Answer:

The time before the ballast hits the ground is 1.186918s

Explanation:

We assume that the upward direction is positive.

The bag released from the ballon is at rest. The velocity of the bag equals the velocity of the balloon.

The balloon has a velocity (V0y) = 3m/s

When the bag is released, it has a height of 5.36m

To touch the ground it makes a displacement of y = - 5.36m

The motion of the bag can be described as followed:

y = V0t + 1/2 * at²

t² + (2V0/a)*t - (2y/a) = 0

We can solve this equation for t:

t= ((-2V0/a) ± √((2V0/a)² + 4*(2y/a))) /2

t = -(V0/a) ± √((V0/a)² + (2y/a))

In this equation we can plug the given values ( V0 = 1.3 m/s ; y = 5.36m ; g =-9.8 m/s²)

t = (1.3/-9.8)±√((1.3/-9.8)²+(2*(-5.36)/-9.8))

t = 0.132653 ± 1.054266

t= 1.186918

The time before the ballast hits the ground is 1.186918s

6 0

3 years ago