How to Find Heat Absorbed and Entropy Change in a Copper Block?

AI Thread Summary
To find the energy absorbed as heat for a 2.00 kg copper block heated from 25°C to 100°C, the calculation yields 57,900 J using the formula dQ = cmΔT. For the change in entropy, since copper is not an ideal gas and the temperature is not constant, the appropriate approach involves integrating the heat capacity over the temperature range. The formula dS = cm ln(Tf/Ti) can be applied, ensuring temperatures are in Kelvin. This method provides a valid way to calculate the entropy change for the copper block. Understanding these thermodynamic principles is crucial for solving related problems.
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Homework Statement


Find (a) the energy absorbed as heat and (b) the change in entropy of a 2.00 kg block of copper whose
temperature is increased reversibly from 25°C to 100°C. The specific heat of copper is 386 J/kg K .


Homework Equations



dS=dQ/T
dS=nRln(final volume/initial volume) **for an ideal gas**

The Attempt at a Solution

\

Well I got the first part. The energy is just
dQ=cmdT
which is (386)(2)(75K)= 57900J

however, when i got to the second part, I didn't know what to do since the copper isn't an ideal gas, and it has a non-constant temperature so I can't use the first equation. Any help would be appreciated!
 
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Entropy Thermodynamics problem

Homework Statement



Find (a) the energy absorbed as heat and (b) the change in entropy of a 2.00 kg block of copper whose
temperature is increased reversibly from 25°C to 100°C. The specific heat of copper is 386 J/kg K .

Homework Equations



dS=dQ/T
dS=nr ln(vf/vi) **for an ideal gas**

The Attempt at a Solution



The first part was relatively easy. I just used the equation:
dQ=cmdT
so, dQ=(386)(2)(75)=57900 J

But for the 2nd part, I don't know what to do because copper is not an ideal gas, ruling out the 2nd equation, and the temperature is not constant, meaning I cannot use the first equation. Any help would be great!
 


dS = dQ/T
dQ = cmdT
dS = cmdT/T
Integrate both sides to get delta(S) = cm*ln(Tf/Ti); be sure to use Kelvin.
Its been a while since I've done thermodynamics, so I might be wrong...though I only applied a bit of mathematical procedure, so it seems as though it would be fine.
 


sounds right to me... thanks for the help!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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