Change in internal energy when water is heated from 0 to 4c

[randomgamernerd]

Homework Statement

: [/B]Find the change in internal energy of 2kg water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200J/Kg and its densities at 0°C and 4°C are 999.9 kg/m³ and 1000kg/m³ respectively.
Atm pressure=10⁵Pa

Homework Equations

:ΔU= Q-W
W=PΔV
M/V=D[/B]

The Attempt at a Solution

:[/B]
ΔW = PΔV
P=10⁵Pa

ΔV= M(1/D₂-1/D₁)
Plugging the values we get W=-0.02(Volume is decreasing...so definitel W should be negative)
Q=m.s.ΔT
= (2)(4200)(4)=33600
So
ΔU= 33600- (-0.02)
= 33600 + 0.02 J
But the answer is given
33600- 0.02

 

[TSny]

Your answer of (33600 + 0.02) J looks correct to me.

 

[randomgamernerd]

Your answer of (33600 + 0.02) J looks correct to me.

then it must be a printing error...right?

 

[mjc123]

Since you are only given the heat capacity to 2 sig figs, correcting the answer by 0.02 seems pointless. But perhaps it's worth demonstrating that it's insignificant?

 

[randomgamernerd]

Since you are only given the heat capacity to 2 sig figs, correcting the answer by 0.02 seems pointless. But perhaps it's worth demonstrating that it's insignificant?

I'm sorry I didnt get you.

 

[mjc123]

Unless otherwise indicated, 4200 means "between 4150 and 4250". Which means Q is between 33200 and 34000 J. With this imprecision, it is meaningless to talk of a correction of 0.02 J.

 

[randomgamernerd]

Unless otherwise indicated, 4200 means "between 4150 and 4250". Which means Q is between 33200 and 34000 J. With this imprecision, it is meaningless to talk of a correction of 0.02 J.

oh..ok..I thought you were giving a hint on how to approach the sum

 

[Chestermiller]

My take on this problem is different. If the initial state is 2 kg water at 0 C and 1 atm, and the final state is 2 kg of water of 4 C and 1 atm, then the path between these two states is irrelevant (since U is a function only of state). Assuming that the specific heat capacity given in the problem statement is $C_p$, the change in enthalpy between the two states is given by:$$\Delta H=mC_p\Delta T$$The change in internal energy between the two states follows from the definition of the change in enthalpy: $$\Delta H=\Delta U+\Delta (PV)=\Delta U+P_{atm}\Delta V$$So, combining these equations, we get $$\Delta U=mC_p\Delta T -P_{atm}\Delta V$$My point is, there is no need to directly apply the first law of thermodynamics to solve this problem.