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 dа ta:

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