# Conservation of energy of a block of ice

• Lunar Guy
In summary, a 2.5 kg block of ice slides across a level floor at a speed of 5.7 m/s. If 3.3 × 10^5 J are required to melt 1.0 kg of ice, then 1.3 × 10^-4 kg of ice melts.
Lunar Guy
Well... It is first first time on the Physics Forums, and I'm stuck on this problem:

1. A 2.5 kg block of ice at a temperature of 0.0 °C and an initial speed of 5.7 m/s slides across a level floor. if 3.3 × 10^5 J are required to melt 1.0 kg of ice, how much ice melts, assuming that the initial kinetic energy of the ice block is entirely converted to the ice's internal energy?

2. ∆PE + ∆KE +∆U = 0
(the change in potential energy + the change in kinetic energy + the change in internal energy = 0)
PE = mgh
KE = ½mv²

3. Well... I really have no idea how to solve this one...

But it sounds to me that KE = ½(2.5 kg)(5.7 m/s)² somewhere in there... But I really don't know what I am supposed to solve for (I think it is the mass of how much ice melts). I don't know how to integrate 3.3 × 10^5 J into an equation. If I could get a equation to start off from, I could do the rest. Thanks.

Last edited:
Are you sure the problem doesn't give you a *final* velocity? The basic concept is that while the block is sliding across the floor, friction does negative work on it, reducing its kinetic energy. The kinetic energy lost heats up the block (i.e. it is converted into internal energy). If you knew the final speed, you'd know the change in kinetic energy. From that, you could calculate how much ice melts.

No final velocity is given... Is there a way I can find it from that information?

EDIT: 3.3 × 10^5 J is circled in my textbook for some reason...

Last edited:
Yeah, maybe we are meant to interpret "the initial kinetic energy is ENTIRELY converted..." as meaning all of it. In that case, the final kinetic energy would be zero.

All right, I'll attempt to work off of that. :)

Here we go:

PEi + KEi +Ui = PEf + KEf + Uf

PEi & PEf = 0

KEf = 0

KEi +Ui = Uf

KEi = Uf - Ui <--- (∆U)

½mv² = ∆U

½(2.5 kg)(5.7 m/s)² = ∆U

42 J = ∆U

3.3 × 10^5 J/1.0 kg = 42 J/m

m(3.3 × 10^5 J) = (42 J)(1.0 kg)

m = 42 J(kg)/3.3 × 10^5 J

m = 1.3 × 10^-4 kg

1.3 × 10^-4 kg of ice melts...

That's not much, isn't it? I hope this is correct...

Last edited:
Does it seem that I got the right answer?

## 1. What is conservation of energy?

Conservation of energy is the principle that energy cannot be created or destroyed, but can only be transformed from one form to another.

## 2. How does conservation of energy apply to a block of ice?

A block of ice has potential energy due to its position in the gravitational field and kinetic energy due to its motion. As the ice melts, its potential energy decreases but its kinetic energy increases, illustrating the conservation of energy.

## 3. Can the energy of a block of ice change forms?

Yes, the energy of a block of ice can change forms. As the ice melts, its potential energy is converted into kinetic energy, and then into thermal energy as the ice warms up.

## 4. How does the conservation of energy affect the melting of a block of ice?

The conservation of energy states that the total energy of a system remains constant. As a block of ice melts, the potential energy decreases but the total energy of the system remains the same.

## 5. How does the conservation of energy apply to the environment?

The conservation of energy is a fundamental principle of the natural world and applies to all systems, including the environment. Energy cannot be created or destroyed, so it is important to use it efficiently and sustainably to minimize its impact on the environment.

• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
3K
• Introductory Physics Homework Help
Replies
19
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
24
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
939
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
262