Question

How many kilojoules are needed to convert 77.5 grams of ice at -8.00°C to water at 95.0°C?


Ahh, Thank you!!! ( ;-;)​

Answer 1

q=mcθ
m= 0.0775 kg
c= 4184 J·kg−1·K−1.
θ= T(final) - T(initial), 95-(-8)

Q=33398.78 J so kJ divide 1000

= 33.39878 kJ