50 points!

how much heat energy is needed to convert 10.75g of ice at -22 celsius to steam at 118 celsius?

A heat energy equal to 6300 J is required to heat 10.75 g of ice to steam.

Explanation:

Given:

Amount of transferred energy = Q =?

Mass of ice (water) = m = 10.75 g

Initial temperature of ice =

Final temperature of steam =

Formula for Heat capacity is given by

Q = m×c×Δt ........................................(1)

where:

Q = Heat capacity of the substance (in J)

m=mass of the substance being heated in grams(g)

c = the specific heat of the substance in J/(g.°C)

Δt = Change in temperature (in °C)

Δt = (Final temperature - Initial temperature) =

Specific heat of water is c = 4.186 J /g. °C

Substituting these in equation (1), we get

Q = m×c×Δt =

Required heat energy is equal to 6299.93 Joules ≅ 6300 Joules.

Visible light is located in the middle of the spectrum.

The reaction 2kalsi3o8 + 2(h+ + hco3-) + h2o  →  al2si2o5(oh)4 + 2k+ + 2hco3- + 4sio2 shows that feldspar kalsi3o8 is chemically altered by hydrolysis to form chalky-looking clay mineral kaolinite al2si2o5(oh)4. the chemical breakdown of potassium feldspar due to exposure to acidic water results to the mineral binding with the water molecule to become kaolinite with silica and potassium and bicarbonate ions in solution.