Question

Based on the thermodynamic properties provided for water, determine the amount of energy needed for 283 g of water to go from 89.5 °C to 161 °C. Property Value Units Melting point 0.0 °C Boiling point 100.0 °C ΔHfus 6.01 kJ/mol ΔHvap 40.67 kJ/mol cp (s) 37.1 J/mol · °C cp (l) 75.3 J/mol · °C cp (g) 33.6 J/mol · °C

Name: based on the thermodynamic properties provided for water determine the amount of energy needed for 283 g of water to go from 895 c to 161 c property value units melting point 00 c boiling po 97052
Uploaded: 2022-08-21T20:05:47-08:00
Duration: 11 min 16 s
Channel: Susan Hallstrom
Description: based on the thermodynamic properties provided for water determine the amount of energy needed for 283 g of water to go from 895 c to 161 c property value units melting point 00 c boiling po 97052

          Based on the thermodynamic properties provided for water,
determine the amount of energy needed for 283  g of water
to go from 89.5 °C to 161 °C. 
Property
Value
Units
Melting point
0.0
°C
Boiling point
100.0
°C
ΔHfus
6.01
kJ/mol
ΔHvap
40.67
kJ/mol
cp (s)
37.1
J/mol · °C
cp (l)
75.3
J/mol · °C
cp (g)
33.6
J/mol · °C

Added by Toni A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Susan Hallstrom

08/21/2022

Step 1

The molar mass of water (H₂O) is approximately 18.015 g/mol. \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{283 \, \text{g}}{18.015 \, \text{g/mol}} \approx 15.71 \, \text{mol} \] Show more…

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Based on the thermodynamic properties provided for water, determine the amount of energy needed for 283 g of water to go from 89.5 °C to 161 °C. Property Value Units Melting point 0.0 °C Boiling point 100.0 °C ΔHfus 6.01 kJ/mol ΔHvap 40.67 kJ/mol cp (s) 37.1 J/mol · °C cp (l) 75.3 J/mol · °C cp (g) 33.6 J/mol · °C