Question

How many joules of heat are absorbed to raise the temperature of 650 grams of water from 5.00c to it's boiling point, 100c

Answer 1

Answer : The amount of heat absorbed are, 258485.5 J

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat gained  = ?

m = mass of water = 650 g

c = specific heat of water = $4.186J/g^oC$      

$\Delta T=\text{Change in temperature}$ 

$T_{final}$ = final temperature = $100^oC$

$T_{initial}$ = initial temperature = $5^oC$

Now put all the given values in the above formula, we get the final temperature of copper.

$Q=650g\times 4.186J/g^oC\times (100-5)^oC$

$Q=258485.5J$

Therefore, the amount of heat absorbed are, 258485.5 J