Answer to Question #108240 in Molecular Physics | Thermodynamics for Shane Coutain

Question

energy required to change 10.0g of ice at -10.0°c to 10.0g of steam at 110.0°c

Answer 1

$q=q_1+q_2+q_3+q_4+q_5$

$q_1=c_{ice}m\Delta t=2140\cdot 0.01\cdot 10=214 J$

$q_2=\lambda_{ice}m=332400\cdot0.01=3324J$

$q_3=c_{water}m\Delta t=4200\cdot 0.01\cdot100=4200J$

$q_4=L_{water}m=2300000\cdot 0.01=23000J$

$q_5=c_{vapor}m=2130\cdot0.01=21.3J$

$q=214+3324+4200+23000+21.3=30759.3J$