Question

How much energy is required to heat 70 g of water at 20°C to boiling

Answer 1

Answer:

$Q=23,430J$

Explanation:

Hello,

In this case, since we compute the required energy via:

$Q=mC\Delta T$

Whereas m is the mass which here is 70 g, C the specific heat which for water is 4.184 J/(g°C) and ΔT is the temperature difference which is:

$\Delta T=100-20=80\°C$

Therefore, the energy turns out:

$Q=70g*4.184\frac{J}{g\°C}*80\°C\\ \\Q=23,430J$

Best regards.