Quantity of heat, phase change of ice to water

[Tafe]

Homework Statement

A 1x10⁶ kg piece of ice is placed into a lake. How much heat is taken from the lake to raise the temperature of the ice from 0 °C to 1x10^-20 °C? How much volume does the lake increase by?
Latent heat for water is 334x10³ J/kg

Homework Equations

Found in my textbook,
c_ice = 2100 J/kg*K
Q = m c ΔT
Q = mL_f
ρ_water = 1.0x10³ kg/m3

The Attempt at a Solution

First I found how much heat is required to raise the temperature of the ice from 0 °C to 1x10^-20 °C.

Q = mcΔT
Q = (1x10⁶ kg)(2100 J/kg*K)(1x10^-20 °C - 0 °C)
Q = 2.1x10^-11 J

Then I found how much ice would melt with that amount of heat.

Q = mL_f
m = Q / L_f
m = 2.1x10^-11 J / 334x10³ J/kg
m = 6.287x10^-17 kg

Finally, I found the volume of the water, or ice that melted.

ρ = m / V
V = m / ρ
V = 6.287x10^-17 kg / 1.0x10³ kg/m3
V = 6.287x10^-20 m³

My understanding is the adding heat to ice doesn't increase its temperature and instead melts some of it into water that is at the same temperature. So I'm confused about the part asking how much heat is required to raise the temperature of the ice. I don't know if I took the right approach for this problem. Is it because the temperature change is so small?

 

[gneill]

Hi Tafe,

Welcome to Physics Forums.

My understanding is the adding heat to ice doesn't increase its temperature and instead melts some of it into water that is at the same temperature. So I'm confused about the part asking how much heat is required to raise the temperature of the ice. I don't know if I took the right approach for this problem. Is it because the temperature change is so small?

Yes, the temperature change they give is negligible and really quite indistinguishable from zero. I think the idea they want to convey is that even to raise the temperature of the ice by that insignificant amount you first have to melt all of it. So you need to convert the ice to liquid water as the first step. Only then can you raise the temperature of the water by the given amount.

 

[Tafe]

I think the idea they want to convey is that even to raise the temperature of the ice by that insignificant amount you first have to melt all of it. So you need to convert the ice to liquid water as the first step. Only then can you raise the temperature of the water by the given amount.

It makes so much more sense now! I think I got it.

Heat required in phase change, ice at 0 °C to water at 0 °C:
Q = mL_f
Q = (1x10⁶ kg)(334x10³ J/kg)
Q = 3.34x10¹¹ J

Heat required in heating water at 0 °C to 1x10^-20 °C:
Q = mcΔT
Q = (1x10⁶ kg)(4190 J/kg*K)(1x10^-20 °C - 0 °C)
Q = 4.19x10^-11 J

The sum of these amounts is the total heat required to raise the temperature of the ice. The volume increase of the lake is the volume of the the entire piece of ice melted.

 

[gneill]

The energy calculations look good.

Shouldn't you give a numerical value for the lake's increase in volume?

 

[Tafe]

Shouldn't you give a numerical value for the lake's increase in volume?

The ice is now water, so
ρ_water = 1.0x10³ kg/m³

ρ = m / V
V = m / ρ
V = (1x10⁶ kg) / (1.0x10³ kg/m³)
V = 1000 m³

 

[gneill]

Looks good.

 

[Tafe]

Thank you so much gneill. :)

 

[gneill]

You're welcome!