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?

I will report if you say the answer is fhfddhjfshfjdfhdjefnhkjfsdewhfid

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=(Q)/(mc)`

`\\ \rm\longmapsto \Delta T=(12552)/(50(4.186))`

`\\ \rm\longmapsto \Delta T=(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`