Question

A 50 ml sample of water at 5 degrees Celsius has 300 calories of heat added. What is the final temperature of the water?
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Answer 1

First we need to find Total change in temperature

• Heat=Q=300C=12552J
• Mass=50ml=50g
• Specific heat capacity=c=4.186J/g°C[/tex]

Now

$\\ \rm\longmapsto Q=mc\Delta T$

$\\ \rm\longmapsto \Delta T=\dfrac{Q}{mc}$

$\\ \rm\longmapsto \Delta T=\dfrac{12552}{50(4.186)}$

$\\ \rm\longmapsto \Delta T=\dfrac{12552}{209.3}$

$\\ \rm\longmapsto \Delta T=59.9\approx 60°C$

Now

$\\ \rm\longmapsto \Delta T=T_2-T_1$

$\\ \rm\longmapsto 60=T_2-5$

$\\ \rm\longmapsto T_2=60+5=65°C$