(i) Heat added
Q=mCPdtQ = m C_PdtQ=mCPdt
Q=m×Cp(2.5+40÷T+20)×dtQ=m \times Cp(2.5+40÷T+20)\times dtQ=m×Cp(2.5+40÷T+20)×dt
Q=1×4.187(2.5+40÷400+20)×400=37.85kJQ= 1 \times 4.187(2.5+40÷400+20)\times 400=37.85kJQ=1×4.187(2.5+40÷400+20)×400=37.85kJ
(ii) Work done
W=p(V2−V1)=200000(1.8−1)=160kJW = p (V_2-V_1)=200000(1.8-1)=160 kJW=p(V2−V1)=200000(1.8−1)=160kJ
(iii) Change in internal Energy:
U=W−Q=160−37.85=122.15kJU = W - Q =160-37.85 =122.15 kJU=W−Q=160−37.85=122.15kJ
(iv) Change in Enthalpy:
dH=dQ=Q=37.85kJdH = dQ =Q = 37.85 kJdH=dQ=Q=37.85kJ
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