For a particular reaction, the system absorb $5 k J$ of heat and does $2 k J$ of work on its surrounding. What is the change in internal energy?

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Solution

Given that the work is done on the surrounding, i.e., the work is done by the system and the heat is absorbed by the system.
Therefore,
$q = 5 k J$
$w = - 2 k J$
Now from first law of thermodynamics,
$Δ U = q + w$
$\Rightarrow Δ U = 5 + (- 2) = 3 k J$
Hence the change in internal energy is $3 k J$ .

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For a particular reaction, the system absorb 5 kJ of heat and does 2kJ of work on its surrounding. What is the change in internal energy?

Given that the work is done on the surrounding, i.e., the work is done by the system and the heat is absorbed by the system.Therefore,q=5kJw=−2kJNow from first law of thermodynamics,ΔU=q+w⇒ΔU=5+(−2)=3kJHence the change in internal energy is 3kJ.

For a particular reaction, the system absorb $5 k J$ of heat and does $2 k J$ of work on its surrounding. What is the change in internal energy?