How many grams of ice at -15.5?C can be completely converted to liquid at 23.6?C
ID: 776236 • Letter: H
Question
How many grams of ice at -15.5?C can be completely converted to liquid at 23.6?C if the available heat for this process is 4.54
Heat, q, is energy transferred between a system and its surroundings. For a process that involves a temperature change where Cs is specific heat and m is mass. Heat can also be transferred at a constant temperature when there is a change in state. For a process that involves a phase change where, n is the number of moles and delta H is the enthalpy of fusion, vaporization, or sublimation. The following table provides the specific heat and enthalpy changes for water and ice. How many grams of ice at -15.5 degree C can be completely converted to liquid at 23.6 degree C if the available heat for this process is 4.54 10^3kJ ? For ice, use a specific heat of 2.01 J/(g. degree C) and delta H fus=6.01kJ/mol.Explanation / Answer
As we know specific heat , c or water is c = 4.18 j/g. oC
Heat needed for 1 gram of ice to get to 0 degree C
q1 = m c delta T
t1 = -15.5 degree C
t 2 = 0 degree C
c = 2.01 J/g . degree C
m = 1 gram
q1 = 1. 2.01
q 1= 1 * 2.1 * [0- (- 15.5)]
q1 = 31.155 J
Next : Calculate the amount of heat needed for one gram of ice to water (heat of fusion)
Delta Hfus = 6.01 kJ / mole
Delta H fus = 6.01 kJ/mole [ 1000 J/ kJ]* [1 mole ice / 18 grams] = 334 J/gram
m = 1 gram
q2 = 1 gram * 334 J / gram = 334 J
Next
Find the amount of heat needed to get one gram up to 23.9 degre C
m = 1 gram
delta t = 22.9 - 0 = 22.9 oC
c = 4.19 J/g* oC
therefore q3 = m.c.delta T
q3 = 1 * 4.19 * 22.9
q3 = 95.95 J
Next
Find the amount of heat in total
Total heat = q1 + q2 + q3 = 31.155 +334 + 95.950 = 461.105 J
then
Find the number of grams that can go through this process.
4.54 x 10^3 kJ = 4540 000 Joules is energy for the process
As calculated above each gram requires 461.105 J
4540 000 J * [1 gram/461.105 J ] = 9845.913 grams
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