Question

A 50.0 g sample of water, initially at 25.0°C, is heated until it starts to boil. How much heat was needed to heat it to 100.0°C? The specific heat capacity of water is 4.184 J/gC.

( I don't know what formula I would use to solve this.)

Answer 1

Step 1: Conversion of water from 25°C to 100°C Mass of water = 50.0 g Heat capcity of water, s=4.184– Amount of heat required, 91 = mSAT = 50.0 g x 4.184 x (100°C-25°C) mol °C = 15690 5x hy = 15.69 kJ - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step 2: conversion of water from 100°C to 100°C: Heat of vaporization of water = 40.7 kJ/mol Amount of heat required, 4z = 50.0 g * 18.02% 1 mol 1 mol 40.7 kJ = 112.93 kJ Total amount of heat required, q = 91 +92 =15.69 kJ +112.93 kJ = 128.62 kJ = 129 kJ