100^{circ} C temperature wala ek garam tambe | vedclass

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(C) Entropy mein parivartan ka sutra $\Delta S = \int \frac{dQ}{T}$ hai.Tambe ke liye,entropy mein parivartan $\Delta S_{Cu} = \int_{T_i}^{T_f} \frac{

Vedclass · Accepted Answer

tamba entropy khota hai aur talab is matra se adhik entropy prapt karta hai

Vedclass · Answer

(C) Entropy mein parivartan ka sutra $\Delta S = \int \frac{dQ}{T}$ hai.Tambe ke liye,entropy mein parivartan $\Delta S_{Cu} = \int_{T_i}^{T_f} \frac{mc dT}{T} = mc \ln(\frac{T_f}{T_i})$ hai,jo ki negative hai (tamba entropy khota hai).Talab ke liye,usne $Q = mc(T_i - T_f)$ ushma prapt ki hai. Talab ka taapman $T_{pond} = 30^{\circ} C$ sthir rehta hai,isliye talab dwara prapt entropy $\Delta S_{pond} = \frac{Q}{T_{pond}} = \frac{mc(100 - 30)}{303.15}$ hai.Kyuki yeh ek anutkramniya (irreversible) prakriya hai,isliye kul entropy mein vriddhi honi chahiye $(\Delta S_{total} > 0)$.Isliye,$\Delta S_{pond} + \Delta S_{Cu} > 0$,jiska arth hai ki $\Delta S_{pond} > |\Delta S_{Cu}|$.Atah,talab tambe dwara khoyi gayi entropy se adhik entropy prapt karta hai.

$100^{\circ} C$ temperature wala ek garam tambe ka tukda $30^{\circ} C$ temperature wale talab mein dala jata hai. Tamba thanda hokar $30^{\circ} C$ par aa jata hai,jabki talab bahut bada hone ke karan apne prarambhik taapman par bana rehta hai. Tab,

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