What kind of friction is occurring between a pencil and desktop when you flick the pencil?

Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,

Nitella [24]

Answer: The amount of heat needed is 29000 Cal.

Explanation:

The process involved in this problem are:

$(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_{p,s}\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 200 g

$T_2$ = final temperature = $0^oC$

$T_1$ = initial temperature = $-30^oC$

Putting all the values in above equation, we get:

$q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal$

For process 2:

$q_2=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 200 g

$L_f$ = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

$q_2=200g\times 80Cal/g=16000Cal$

For process 3:

$q_3=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water = 200 g

$T_2$ = final temperature = $50^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal$

Calculating the total heat absorbed, we get:

$Q=q_1+q_2+q_3$

$Q=3000+16000+10000=29000Cal$

Hence, the amount of heat needed is 29000 Cal.

7 0

3 years ago

What is the color of visible light of the lowest frequencies? Of the highest frequencies?

Vikentia [17]

highest - violet

lowest - red

3 0

3 years ago

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g A circular loop of wire with radius 3 m is flat on the xy plane in a magnetic field which is pointed in the z-direction and ha

Phantasy [73]

Answer:

emf=-0.035V

Explanation:

this problem can be solved with the Lenz's law:

$emf=-\frac{\Delta \Phi_{B}}{\Delta t}$

where PhiB is the magnetic flux. In this case we have

$\Delta \Phi_{B}=B(A_1-A_2)$

due to the magnetic field is constant. A is the area of the circular loop. Hence

$\Delta \Phi_{B}=(5*10^{-3}T)(\pi(3m)^2-\pi(1.5m)^2)=0.106Tm^2$

Finally

$emf=-\frac{0.106Tm^2}{3s}=-0.035V$

HOPE THIS HELPS!!

7 0

3 years ago

Name one contact force and one force that acts through a force field

nirvana33 [79]

A contact force is a type of force which act on an object by coming in contact with the object. Examples of contact force that acts through a force field are: applied force, frictional force, air resistance force, tension, spring force, etc.
Examples of forces that act through a force field are gravitational force, electromagnetic force, the weak interaction and the strong interaction.

5 0

4 years ago

What's on the other side of a black hole

bearhunter [10]

Not sure, nobody knows hopefully soon we will know

6 0

4 years ago

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