Deepak Patra
Last Activity: 10 Years ago
Sol. Mass = 10g = 0.01kg.
P = 10^5Pa
dQ = Q base h2o 0° + Q base H2o – steam
= 0.01 × 4200 × 100 + 0.01 × 2.5 × 10^6 = 4200 + 25000 = 29200
dW = P × ∆V
∆ = 0.01/0.6 – 0.01/1000 = 0.01699
dW = P∆V = 0.01699 × 10^5 1699J
dQ = dW + dU or dU = dQ – dW = 29200 – 1699 = 27501 = 2.75 × 10^4 J