Determine the heat energy needed to raise the temperature of 120 grams of ice at -5 to steam at 115°

Q = 30355.2 J

Explanation:

Given data:

Mass of ice = 120 g

Initial temperature = -5°C

Final temperature = 115°C

Energy required = ?

Solution:

Specific heat capacity of ice is = 2.108 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT = T2 -T1

ΔT = 115 - (-5°C)

ΔT = 120 °C

Q = 120 g × 2.108 j/g.°C × 120 °C

Q = 30355.2 J

they are all solids.