Question

50 grams of ice-cold water (0 °C) is heated to boiling (100 °C). The specific heat capacity of water is 4.18 J /g°C. How much energy did the water absorb?

Answer 1

$Q= m.c.∆T$

Q=heat absorbed

$m=mass=50grams$

$c$ = specific heat capacity $=418J/g°C$

$∆T=100-0=100°C$

$Q=50×4.18×(100-0)=20900J$