Calculate the heat energy required to change 0.1kg of ice at 0^oC to water boiling at 100^oC. (Specific heat capacity of water = 4200 J kg^-1K^-1) (Specific latent heat of fusion of ice = 336,000 J kg^-1)

Kayden

5 years ago

H = mL + mc

H = 0.1×336000 + 0.1×100×4200

H = 33600+42000

H = 75600

Abisola_16

3 years ago

H = mL + mc * change in tita
H = 0.1*336000 + 0.1*100*4200
H = 33600 + 42000
H = 75600J

Xofficial

4 years ago

H = mL + mc * change in tita

H = 0.1×336000 + 0.1×100×4200

H = 33600+42000

H = 75600

Gobbler

9 years ago

Please explain

FGCPOR17576703

3 years ago

Guyy, its 336000 they gave and people are solving with 33600. how na