Thermal energy in boiling water

[mshah3]

Homework Statement

140 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are poured into an aluminum pan whose mass is 375 grams and initial temperature 25° C (the heat capacity of aluminum is 0.9 J/gram/K).
(a) After a short time, what is the temperature of the water?
Tfinal = 72.6499° C
(b) What simplifying assumptions did you have to make?
Energy transfer between the system (water plus pan) and the surroundings was negligible during this time.
The thermal energy of the aluminum doesn't change.
The thermal energy of the water doesn't change.
The heat capacities for both water and aluminum hardly change with temperature in this temperature range.

(c) Next you place the pan on a hot electric stove. While the stove is heating the pan, you use a beater to stir the water, doing 1300 J of work, and the temperature of the water and pan increases to 79.1° C. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan?
Q = -1300J

Homework Equations

thermal energy change of water + thermal energy change of aluminum = 0
thermal energy change = mC(delta T)


E = W + Q
0 = W + Q
Q = -W

The Attempt at a Solution

the answers in red are the ones i have calculated using the above equations.

need someone to confirm that i have done this correctly

only have one submission left : /

 

[learningphysics]

I get the same Tfinal...

"The thermal energy of the aluminum doesn't change.
The thermal energy of the water doesn't change."

but they do change right? but the thermal energy of the whole system doesn't change...

for the last part...
E = W + Q

Q = E - W

you get E from mc*deltaT... you have W = 1300...

I think that's the way to do it...

 

[mshah3]

oh ok ...i understand for the first two parts.
i have a question bout the last part tho in finding Q
not quite sure which mass, specific heat, or delta T to use
i was thinkin to use just the values for aluminum

so it would be:

E= (375)(0.9)(79.1-72.6499) = 2176.9087

Q= E-W = 2176.9087-1300 = 876.9087

did i do this correctly?

 

[learningphysics]

oh ok ...i understand for the first two parts.
i have a question bout the last part tho in finding Q
not quite sure which mass, specific heat, or delta T to use
i was thinkin to use just the values for aluminum

so it would be:

E= (375)(0.9)(79.1-72.6499) = 2176.9087

No, you need to also add the mc*deltat of the water.

Q= E-W = 2176.9087-1300 = 876.9087

did i do this correctly?

fix the E to the sum of the mc*deltat of pan + mc*delta of water... then everything will be good.

 

[mshah3]

oh ok...so does my delta t remain the same whether I am calculatin for water or for pan?

delta t= tfinal-tinitial = 79.1-72.6499 = 6.4501

 

[learningphysics]

oh ok...so does my delta t remain the same whether I am calculatin for water or for pan?

delta t= tfinal-tinitial = 79.1-72.6499 = 6.4501

yes.

 

[mshah3]

ok awesome.
i got a final value of 4669.5675 J
can u confirm this?

 

[learningphysics]

ok awesome.
i got a final value of 4669.5675 J
can u confirm this?

yes, that's what I get.