the sum of the heats gained is zero (some will be lost).
solve for Tf.
The 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)
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
(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
320g x 4.186j/g-C x T
+ 16700j = 0
1339.52j/C x T = -5397.8j
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(ice) = mass? x (334000j/kg*1kg/1000g)
Energy balance tells us:
DeltaH(hotwater)+DeltaH(ice) = 0
-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
33.84g of ice melted after adding 30g of hot water to the system, but the system is still at 0deg celcius.
physics(heat) - a mixture of 250g of water and 200g of ice at 0deg C is kept in ...
PHYS101 - In a calorimeter can(water is melted above?
physics - A piece of metal of mass 112g is heated 100 degree C.,and dropped into...
physic - a resistor has a voltage of 6.65 V and a current of 4.45 A. it is ...
chemistry - In the following experiment, a coffee-cup calorimeter containing ...
Chemistry - A 5.00g sample of octane is burned in a bomb calorimeter containing ...
Chemistry 2 - I have to calculate the Ccal using the following info: Initial ...
Chemistry - A calorimeter contains 50.0 mL of water; both are at 21.2 degrees C...
Chemistry - General - 1. Consider a calorimeter into which 135g of water at 45.3...
CHEMISTRY - A student calibrates his/her calorimeter. S/he performs part 1 as ...