Question

How many kJ are required to heat 45.0 g of H2O at 25.0°C and then boil it all away?

Answer 1

Answer : The amount of heat required is, 14.12 KJ

Solution :

Formula used :

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

where,

Q = heat required = ?

m = mass of water = 45 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 = $25^oC$

Now put all the given values in the above formula, we get the amount of heat required.

$Q=45g\times 4.186J/g^oC\times (100-25)^oC$

$Q=14127.75J=14.12KJ$

Therefore, the amount of heat required is, 14.12 KJ