Heat transfer calculation problem

AI Thread Summary
The discussion focuses on a heat transfer calculation involving water and ice, where the goal is to determine the mass of ice needed to reach a final temperature of 25.0°C. The user initially applies the heat transfer equations incorrectly, leading to an incorrect mass calculation of ice. Key points include the need to account for the phase change of ice as it warms to 0°C and then melts, which requires additional energy calculations. The user realizes that the specific heat of water must be considered after the ice melts, adjusting their approach accordingly. Ultimately, the correct method involves equating the heat lost by the water to the heat gained by the ice and the resulting water.
KurtWagner
Messages
43
Reaction score
1
I must be doing something wrong here but I cannot figure out what it is1. Homework Statement
An insulated beaker with negligible mass contains a mass of 0.350kg of water at a temperature of 70.1∘C.
How many kilograms of ice at a temperature of − 11.6∘C must be dropped in the water to make the final temperature of the system 25.0∘C?

Take the specific heat for water to be 4190J/(kg⋅K) , the specific heat for ice to be 2100J/(kg⋅K) , and the heat of fusion for water to be 334kJ/kg .

Homework Equations



Q = mcT
Q = mcT + Lfm

The Attempt at a Solution



So basically I was just doing...

mwcw ΔTw = mi [(ci ΔTi) + Lf i]

which gives
ΔTw = 70.1∘C. - 25∘C = 45.1K
ΔTi = − 11.6∘C - 25∘C = 36.6K

0.350kg * 4190J/(kg⋅K) * (45.1K) = mi [(2100J/(kg⋅K)*(36.6K)] +334kJ/kg]
66139.15J = 257140J/kg * mi
mi = 0.161kg

which does not turn out to be the right answer (I do not have access to the right answer).
Can anyone tell me what mistake I am making.
Any help would be greatly appreciated.
 
Physics news on Phys.org
What happens to the ice when it reaches a temperature of 0 C? Is your equation equating the heat lost by the water in the beaker to the heat gained by the ice dropped into the beaker correct? You should first write out what happens to the ice after it is dropped into the beaker without applying any numbers.
 
  • Like
Likes KurtWagner
Q = mi [(ci ΔTi) + Lf i]

Would this equation not mean that each kilo of ice absorbs the heat needed to change its state and also to change its temp?
for each kilo 2100 Joules per degree change in temp but also the 334kJ to change its state?

Am i right in assuming that each kilo of ice changes through -11.6∘C to 0∘C, taking 24360J , changing state for 334000J, then continuing to change from 0∘C to 25∘C taking a further 52500J. Total = 410860J/kg

Whoops, that's different from above, but It is what I used in my calculations
 
oh... It becomes water and has a different specific heat after 0∘C... hmmmm. Ill try that
 
That worked.

Qwater = Qice + Qicewater
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top