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

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.