Question

If 11.9 kJ are used to heat a sample of water the temperature increases from 20.0°C to

Answer 1

Answer:

$m=4.51g$

Explanation:

Hello!

In this case, since the energy involved during a heating process is shown below:

$Q=mCp\Delta T$

Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:

$m=\frac{Q}{Cp\Delta T}$

Thus, by plugging in, we obtain:

$m=\frac{11900J}{4.184\frac{J}{g\°C}(650\°C-20.0\°C)}\\\\m=4.51g$

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