Question

Water (2230 g ) is heated until it just begins to boil. If the water absorbs 5.73×105 J of heat in the process, what was the initial temperature of the water?

Answer 1

Answer: $38.5^0C$

Explanation:

To calculate the initial temperature of the water:

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

where,

q = heat absorbed = $5.73\times 10^5J$

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 2230 g

$T_{final}$ = final temperature of water = $100^0C$

$T_{initial}$ = initial temperature of metal = ?

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

$5.73\times 10^5J=2230g\times 4.18J/g^oC\times (100-T_i)^0C$

$(100-T_i)=61.5$

$T_i=100-61.5)=38.5^0C$

Thus, the initial temperature of the water is $38.5^0C$