Heat transfer calculation problem

In summary, after dropping 0.350kg of ice into a beaker of water at a temperature of 70.1∘C, the final temperature of the system will be 25.0∘C. This can be explained by the addition of the heat gained by the ice dropped into the system and the heat lost by the water in the beaker.
  • #1
KurtWagner
43
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.
 
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  • #2
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.
 
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Likes KurtWagner
  • #3
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
 
  • #4
oh... It becomes water and has a different specific heat after 0∘C... hmmmm. Ill try that
 
  • #5
That worked.

Qwater = Qice + Qicewater
 

Related to Heat transfer calculation problem

1. What is heat transfer calculation?

Heat transfer calculation is the process of determining the amount of heat that is transferred from one object to another. It involves using mathematical equations and physical principles to calculate the rate of heat transfer.

2. What are the three types of heat transfer?

The three types of heat transfer are conduction, convection, and radiation. Conduction is the transfer of heat through direct contact between two objects. Convection is the transfer of heat through the movement of fluids, such as air or water. Radiation is the transfer of heat through electromagnetic waves.

3. How is heat transfer calculated?

Heat transfer is calculated using the equation Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.

4. What factors affect heat transfer?

The rate of heat transfer can be affected by various factors, including the temperature difference between the two objects, the surface area of contact, the thermal conductivity of the materials, and the distance between the objects.

5. Why is heat transfer calculation important in science?

Heat transfer calculation is important in science because it allows us to understand and predict how heat is transferred in various systems. This is crucial in fields such as thermodynamics, engineering, and meteorology, where heat transfer plays a significant role in the behavior of systems and the environment.

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