How much heat is required fo change 28g of ice at -9.00 C to 28g of steam at 131 C

Answer 1

Thermodynamics.

This is a multi-step process where we have to find the heat to change 28 g of ice at -9ºC to steam at 131ºC. We will need to use the following constants:

C_ice = 2.09 J/g/deg

C_H2O = 4.184 J/g/deg

C_steam = 1.89 J/g/deg

∆H_fusion = 334 J/g

∆H_vaporization = 2260 J/g

Step 1. Find the heat needed to change 28 g of ice at -9ºC to ice at 0ºC.

q = mC∆T where q = heat; m = mass; C = specific heat of ice; ∆T = change in temperature

q = (28 g)(2.09 J/g/deg)(9 deg) = 527 J

Step 2. Fine the heat to melt 28 g of ice at 0ºC (phase change; no change in temperature)

q = m∆H_fusion = (28 g)(334 J/g) = 9352 J

Step 3. Find the heat to raise raise temperature of 28 g H2O from 0º to 100 ºC.

q = mC∆T = (28 g)(4.184 J/g/deg)(100 deg) = 11,715 J

Step 4. Find the heat to change 28 g of H2O to steam at 100ºC (phase change; no change in temperature)

q = m∆Hvaporization = (28 g)(2260 J/g) = 63,280 J

Step 5. Find the heat to raise the temperature of 28 g of steam from 100º to 131ªC.

q = mC∆T = (28 g)(1.89 J/g/deg)(31 deg) = 1641 J

Add up all of the heat: 527 J + 9352 J + 11,715 J + 63,280 J + 1641 J = 86,515 J

So the answer is 86,515 Joules or 86.5 kJ