Investigating Ice Specific Heat of Fusion

[chawki]

Homework Statement

The ice piece of 380 g was taken from the deep-freezer, temperature was -9,6 °C, and well insulated container to the being stone-cold ( 0 °C) water. Inside the Ice piece was small temperature detector.
When temperature of the the ice piece had balanced 0 °C. part of the water was frozen
to the surface. Ice was taken from water and were weighed again, the result gots 403 g.

Homework Equations

a) How much energy was released from the water amount, which froze.
b) Define the ice specific heat capacity.
Ice specific heat of fusion is 334 kJ/kg.

The Attempt at a Solution

a)
we calculate Q ? but with which mass
Q=m*Cp.T

 

[gneill]

The original mass of ice is the sink for the heat, allowing new ice to form. How much new ice formed? How much heat had to be removed from it in order for it to freeze?

 

[chawki]

The original mass of ice is the sink for the heat, allowing new ice to form. How much new ice formed? How much heat had to be removed from it in order for it to freeze?

new ice is 403-380=23g ?

 

[chawki]

Q=m*Cp*T
Q=0.023*4.2*9.6 = 0.92736 J ?

 

[gneill]

Q=m*Cp*T
Q=0.023*4.2*9.6 = 0.92736 J ?

Where did 4.2 come from?

 

[chawki]

I assumed it's the specific heat 4.2Kj/(kg*C)

 

[gneill]

The newly formed ice underwent a phase transition from liquid at zero degrees C to solid at zero degrees C: no temperature change. Use the heat of fusion; that's what it's for.

 

[chawki]

can you please tell me the law for the heat of fusion

 

[gneill]

Take a look at the units attached to it: kJ/kg . Note that, unlike specific heat, there's no °C in the denominator. This suggests that whatever operation takes place (in this case freezing or melting) occurs at a single temperature. And in fact, here we have water at 0C turning to ice at 0C.

So, simply multiply the heat of fusion by the grams of substance that undergoes the change to determine how many Joules of energy will be required or released (melting takes energy input, freezing requires energy to be extracted).

Q = m*H_f

 

[chawki]

Take a look at the units attached to it: kJ/kg . Note that, unlike specific heat, there's no °C in the denominator. This suggests that whatever operation takes place (in this case freezing or melting) occurs at a single temperature. And in fact, here we have water at 0C turning to ice at 0C.

So, simply multiply the heat of fusion by the grams of substance that undergoes the change to determine how many Joules of energy will be required or released (melting takes energy input, freezing requires energy to be extracted).

Q = m*H_f

Ok, so it's
Q=0.38*334
Q=126.92 J

 

[gneill]

The 380 gram piece of ice was already ice, and it didn't melt. So the heat of fusion doesn't apply to its temperature change.

You want to use the newly formed ice, which changed from liquid to solid.

 

[chawki]

ok so it's that 23g ?
Q=0.023*334
Q=7.682 J
( i have to say that the way they explained what they did is still not very clear to me)

 

[gneill]

Watch your powers of ten; That should be kJ.

That's the energy that was extracted from the water in order to turn it to ice. In turn, this energy went into the existing ice to warm it up (from -9.6C).

 

[chawki]

Q = 7682 J ( as i prefer to use Joules)

 

[chawki]

Then how we will find Cp ? from Q=m*Cp*T ?

 

[gneill]

Yes.

 

[chawki]

which mass we use? the initial mass or the mass after after it melted ?

 

[gneill]

Nothing melted. New ice formed. The initial piece of ice warmed up, from -9.6 to 0C. You're interested in the warming of the initial piece of ice due to the heat it obtained from the freezing of the new ice.

 

[chawki]

Nothing melted. New ice formed. The initial piece of ice warmed up, from -9.6 to 0C. You're interested in the warming of the initial piece of ice due to the heat it obtained from the freezing of the new ice.

so it's probably 0.38 Kg
But when we substitute Q as 7.682 kj?