Question

If 200. g of water at 20°C absorbs 41 840 J of energy, what will its final temperature be? (Specific Heat of water is 4.184 J/g*C)

Answer 1

Answer: The final temperature will be $70^0C$

Explanation:

To calculate the specific heat of substance during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed =41840 J

c = specific heat = $4.184J/g^0C$

m = mass of water  = 200 g

$T_{final}$ = final temperature =?

$T_{initial}$= initial temperature = $20^0C$

Now put all the given values in the above formula, we get:

$41840J=200g\times 4.184J/g^0C\times (T_{final}-20)^0C$

$T_{final}=70^0C$

Thus the final temperature will be $70^0C$