Answer to Question #234135 in Molecular Physics | Thermodynamics for Unknown346307

Question

1. In a system, executing a non-flow process, the work and heat per degree

change of temperature are given by

dW

dT

= 200 W-s/°C and dQ

dT = 160 J/°C

What will be the change of internal energy of the system when its temperature changes

from

T1 = 55°C to T2 = 95°C ?

Answer 1

In a system, executing a non flow process.

(NB: non flow process is a process in which no mass transfer takes place between system and surroundings)

From first law of thermodynamics,

Q=W+U

differentiating with respect to time,

$\dfrac{dQ}{dT} = \dfrac{dW}{dT} +\dfrac{ dU}{dT}$

given,

$\dfrac{dQ}{dT} = 200 W.s/°C = 200 J/°C$

$\dfrac{dW}{dT} = 160 J/°C$

so,

$\dfrac{dU}{dT} = 200-160= 40J/°C$

change of internal energy,

$U = \dfrac{dU}{dT}× ∆T= 40 J/°C × (95°C-55°C) = 40J/°C × 40°C = 1600\ J$