Question

I have a physics worksheet to do and I have no idea how to do any of it. It is on heat and energy. Would someone be able to help me please?

Answer 1

I can't read the words on the picture, sorry. The picture is very blurry in parts. Is there something you realize that you don't know that you can ask about?

Answer 2

if I write out one of the problems...would you help me with it?

Answer 3

Possibly! I'd have to see what's involved! :)

Answer 4

a 150 gram piece of ice at -30 degrees celsius is heated until it has been changed completely to steam at 140 degrees celsius. How much heat has been added?

Answer 5

I'm not as good at this stuff.
Do you know about heat capacity and specific heat capacity?

Answer 6

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html#c1

Answer 7

i know about those things...I think I just dont know how to set up the problem

Answer 8

Okay! Do you like using equations that are given to you? The link I sent provides the equation
\(Q=cm\Delta T\)
and
\(c=\ 4.186\ \rm\dfrac{Joule}{gram\dot\ °C}\)
Although I think it should be \(C^\circ\) nowadays.

Answer 9

What that equation tells you is that the left side, \(Q\), a.k.a. energy added not from work, is equal to... Hey, energy is hard to think about. And, even though temperature is also abstract, it's a little easier to think about. So
\(\Delta T=\dfrac{Q}{c\dot\ m}\)

Okay. So the temperature will increase with more heat added.
The temperature won't increase as much for that energy if there is more mass, since it's in the denominator. This makes sense. Conceptually, more mass will take more energy to heat all of it up.
The the temperature won't increase as much with a greater specific heat capacity for the same reason. It makes sense. Conceptually, if the heat capacity is great, it mean it takes A LOT of energy to change the temperature. You see that in this equation.

Answer 10

As ice acquires heat first it goes from -30 to 0, the ice melts and becomes water while staying in a stable temperature then it goes from 0 degrees to 100 degrees. But it is still water and it needs to evaporate to steam. After it does, it gains another 40 degrees.

This was the logic, now comes the calculations.

First step, from -30 to 0.
\[Q=m.c.\Delta T\] c for ice is 9/10
150. 9/10. 30 = 4050

Now the ice will melt since it is 0 degrees celcius.
\[Q=m.L\] L for ice is 334J/g
150.334= 50100

Ice has become water so far. Now it needs to go from 0 to 100 degrees. It cant go to 140 without evaporating so we stop at 100 degrees.

First formula again, c for water is 1

At 100 degrees it will boil. The second formula, L is 2260 for water
150.2260= 339000

Now the rest 40 degrees it will go up.First formula, c for vapor is 0,6.
150.6/10.40=3600

Add up all the heat that is required;

4050+50100+15000+339000+3600= 411750 Joules

I might be wrong, this is all I know and I wanted to try. Feel free to correct.