Question
Asked Nov 7, 2019
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How much heat is required to warm 100.0 g water from 31.8 degrees Celcius to 67.5 degrees Celcius? The specific heat capacity of water is 4.184 J/g K

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Expert Answer

Step 1

The heat required is calculated as,

= mCAT
m 100 g
T1 31.8 °C
304.8 K
T2
67.5 °C 340.5 K
C 4.184 J/g-K
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= mCAT m 100 g T1 31.8 °C 304.8 K T2 67.5 °C 340.5 K C 4.184 J/g-K

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Step 2

Substituting, th...

QmCAT
Qm x C x (T2- Ti)
Q 100 g x 4.184 J/g-K x (340.5 -304.8) K
Q 14936.88 J
Q 14.936 kJ
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QmCAT Qm x C x (T2- Ti) Q 100 g x 4.184 J/g-K x (340.5 -304.8) K Q 14936.88 J Q 14.936 kJ

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