How many Joules are required to change 30.0g ofice at -20°C to steam at 140°C? There are a few steps you should follow using the following given information
• The specific heat of ice is 2.108 J/g*°C
• The specific heat of water is 4.18 J/g*°C
• The specific heat of steam is 2.010 J/g*°C
• Hf for H2O = 334 334 J/g
• Hv for H2O = 2260 J/g

Q = 10118.4 j

Explanation:

Given data:

Mass of ice = 30.0 g

Initial temperature of ice = -20°C

Final temperature = 140°C

Energy required = ?

Solution:

The specific heat 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

ΔT = 140°C -  (-20°C)

ΔT = 160°C

Q = 30.0 g× 2.108 J/g°C ×160°C

Q = 10118.4 j