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

Respuesta :

Answer : The amount of heat absorbed are, 258485.5 J

Solution :

Formula used :

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

where,

Q = heat gained  = ?

m = mass of water = 650 g

c = specific heat of water = [tex]4.186J/g^oC[/tex]      

[tex]\Delta T=\text{Change in temperature}[/tex] 

[tex]T_{final}[/tex] = final temperature = [tex]100^oC[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]5^oC[/tex]

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

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

[tex]Q=258485.5J[/tex]

Therefore, the amount of heat absorbed are, 258485.5 J