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February 5, 2016
Posted by **KABELO** on Thursday, April 29, 2010 at 10:32am.

- phys101 -
**bobpursley**, Thursday, April 29, 2010 at 10:36amthe sum of the heats gained is zero (some will be lost).

masscan*cw*(0-Tf)+200*Cw*(0-Tf)+50*Hfice+ 50*cw*(0-tf)+30*cw*(Tf-90)=0

solve for Tf.

- phys101 -
**Ben**, Thursday, April 29, 2010 at 11:23amThe heat-energy needed to produce any temperature change in the liquid is: DeltaH = mass x heat-capacity x DeltaT

where DeltaT is the change in temperature.

The heat-energy needed to melt the ice is: Hmelting = mass x heat-of-fusion

Lets assume the final temperature is 'T'. Then...

The energy change in the 30g of hot water is:

DeltaH(hotwater) = 30g x (4186j/kg-C*1kg/1000g) x (T-90deg)

The energy change in the 200g of cold water is:

DeltaH(coldwater) = 200g x (4186j/kg-C*1kg/1000g) x (T-0deg)

The energy change in the can is:

DeltaH(can) = 40g x (4186j/kg-C*1kg/1000g) x (T-0deg)

Note: The can should start at the same temperature as the water and ice it contains, so initial temperature=0deg C.

Finally, the energy change in the ice is (assuming the ice melts completely):

DeltaH(ice) = 50g x (334000j/kg*1kg/1000g) + 50g x (4186j/kg-C*1kg/1000g) x (T-0deg)

Because energy is conserved all these changes in energy in the individual components of the system should sum to zero for the whole system:

DeltaH(hotwater)+DeltaH(coldwater)+DeltaH(can)+DeltaH(ice) = 0

So we get:

30g x (4186j/kg-C*1kg/1000g) x (T-90deg)

+

200g x (4186j/kg-C*1kg/1000g) x (T-0deg)

+

40g x (4186j/kg-C*1kg/1000g) x (T-0deg)

+

50g x (334000j/kg*1kg/1000g)

+ 50g x (4186j/kg-C*1kg/1000g) x (T-0deg)

= 0

Now we just have to solve for 'T'.

Simplifying, we get:

30g x 4.186j/g-C x (T-90)

+(200+40+50)g x 4.186j/g-C x (T-0)

+50g x (334j/g) = 0

OR

(30+200+40+50)g x 4.186j/g-C x T

+ 30g x 4.186j/g-C x (-90)

+ 50g x 334j/g = 0

OR

320g x 4.186j/g-C x T

+ -11302.2j

+ 16700j = 0

OR

1339.52j/C x T = -5397.8j

OR

T = -5397.8j/1339.52j/C = -4.029dec C

BUT! This answer is wrong. If we add hot water to cold water and ice the result can never be colder than the starting point (0dec C). This means one of our assumptions was wrong.

We only made two assumptions:

1) the can is the same temperature as the water and ice at the start (0deg)

2) the ice melts completely

The 2nd assumption must be incorrect, so we need to change our calculations accordingly:

If the ice does not melt completely, then there will still be ice left in the system. If ice is left then the final temperature must be 0deg celcius because ice and water at equilibrium must be at 0deg celcius (the melting point, by definition).

A related question then is how much ice melted? This can be solved in a similar way:

DeltaH(hotwater)= 30g x (4186j/kg-C*1kg/1000g) x (0-90deg)

DeltaH(hotwater)= -11302.2j

DeltaH(ice) = mass? x (334000j/kg*1kg/1000g)

Energy balance tells us:

DeltaH(hotwater)+DeltaH(ice) = 0

OR

-11302.2j + mass? x 334000j/kg*1kg/1000g = 0

OR mass? x 334j/g = 11302.2j

OR mass? = 11302.2j / 334j/g

OR mass? = 33.84g

So:

33.84g of ice melted after adding 30g of hot water to the system, but the system is still at 0deg celcius.