Question

5. 5 points Previous Answers Chang7 12.072. WyNote All Your Teacher 2 of 30 team t h e species we weam and are 19 and 18 , respectively. The heat ofis 6.02 km and heat How much heat is needed to convert of vaporization is 40.79 /mol) Your r eferencantly from the cont e nt you to the town and th e care
Can someone help me with this?

Answer 1

Step1: Conversion of ice from -30°C to 0°C Mass of ice= 902 g Heat capcity of ice, s = 2.03 — Amount of heat required, 41 = mSAT = 902 8 * 2.036 (0.0°C-(-30.0°C)) = 54932J --------------------------------------- Step2: Conversion of ice from 0°C to 0°C: Heat of fusion of ice = 334 J/g 334 J Amount of heat required, q2 = 902 g x = 301268J - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step3: Conversion of water from 0°C to 100°C Heat capcity of water, s = 4.18 - Amount of heat required, 93 = mSAT = 902 g * 4.18 ----x(100°C-0°C) = 377036 J -------------------------------------- Step 4: conversion of water from 100°C to 100°C: Heat of vaporization of water = 2260 J/g 2260 J Amount of heat required, 44 = 902 g * --* = 2038520 J - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step 5: conversion of water vapor from 100°C to 124°C Heat capcity of water vapor, s = 1.99 Amount of heat required, qs = mSAT = 902 g * 1.99-6-x(124°C-100°C) = 43080 J Total amount of heat required, q = 91 +92 +93 +94 +95 = 54932 J+301268 J + 377036 J+ 2038520 J + 43080J = 2814836J = 2814.836 kJ = 2815 kJ