P21_015 - entropy is ∆ S = Q/T where Q is the energy it...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
15 .Th ei c ew a rm sto0 C, then melts, and the resulting water warms to the temperature of the lake water, which is 15 C. As the ice warms, the energy it receives as heat when the temperature changes by dT is dQ = mc I dT ,where m is the mass of the ice and c I is the speciFc heat of ice. If T i (= 263 K) is the initial temperature and T f (= 273 K) is the Fnal temperature, then the change in its entropy is S = Z dQ T = mc I Z T f T i dT T = mc I ln T f T i =( 0 . 010 kg)(2220 J / kg · K) ln µ 273 K 263 K =0 . 828 J / K . Melting is an isothermal process. The energy leaving the ice as heat is mL F ,where L F is the heat of fusion for ice. Thus, ∆ S = Q/T = mL F /T =(0 . 010 kg)(333 × 10 3 J / kg) / (273 K) = 12 . 20 J / K. ±or the warming of the water from the melted ice, the change in entropy is S = mc w ln T f T i , where c w is the speciFc heat of water (4190 J / kg · K). Thus, S =(0 . 010 kg)(4190 J / kg · K) ln µ 288 K 273 K =2 . 24 J / K . The total change in entropy for the ice and the water it becomes is S =0 . 828 J / K+12 . 20 J / K+2 . 24 J /
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: entropy is ∆ S = Q/T , where Q is the energy it receives as heat (the negative of the energy it supplies the ice) and T is its temperature. When the ice warms to 0 ◦ C, Q = − mc I ( T f − T i ) = − (0 . 010 kg)(2220 J / kg · K)(10 K) = − 222 J . When the ice melts, Q = − mL F = − (0 . 010 kg)(333 × 10 3 J / kg) = − 3 . 33 × 10 3 J . When the water from the ice warms, Q = − mc w ( T f − T i ) = − (0 . 010 kg)(4190 J / kg · K)(15 K) = − 629 J . The total energy leaving the lake water is Q = − 222 J − 3 . 33 × 10 3 J − 6 . 29 × 10 2 J = − 4 . 18 × 10 3 J. The change in entropy is ∆ S = − 4 . 18 × 10 3 J 288 K = − 14 . 51 J / K . The change in the entropy of the ice-lake system is ∆ S = (15 . 27 − 14 . 51) J / K = 0 . 76 J / K....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online