Question #108240
energy required to change 10.0g of ice at -10.0°c to 10.0g of steam at 110.0°c
1
Expert's answer
2020-04-07T09:05:38-0400

q=q1+q2+q3+q4+q5q=q_1+q_2+q_3+q_4+q_5


q1=cicemΔt=21400.0110=214Jq_1=c_{ice}m\Delta t=2140\cdot 0.01\cdot 10=214 J


q2=λicem=3324000.01=3324Jq_2=\lambda_{ice}m=332400\cdot0.01=3324J


q3=cwatermΔt=42000.01100=4200Jq_3=c_{water}m\Delta t=4200\cdot 0.01\cdot100=4200J


q4=Lwaterm=23000000.01=23000Jq_4=L_{water}m=2300000\cdot 0.01=23000J


q5=cvaporm=21300.01=21.3Jq_5=c_{vapor}m=2130\cdot0.01=21.3J


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








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