Question

at 1 arm, how many energy is required to heat 35.0g of water (s) at -22.0 degree to water (g) at 131.0 degree Celsius?

Answer 1

At 1 atm means, at normal atmospheric pressure.

Step 1: Energy required to heat water(s) at -22 oC to 0 oC.

Q = mc(delta T)

Q = 35.0 x 2.01 x (0 - (-22)) [Specific heat of ice is 2.01 J/g oC]

Q = 1547.7 J

Step 2: Energy required to heat water(s) from 0 oC to water (l) at 0 oC

Q = m x delta Hf [Since, this is a phase conversion from solid to liquid, we use heat of fusion (Hf)]

Q = 35.0 x 334 [ Heat of fusion of water is 334 J/g]

Q = 11690 J

Step 3: Energy required to heat water(l) at 0 oC to water(l) at 100 oC

Q = mc(delta T)

Q = 35.0x 4.18 x (100 - 0)   [Specific heat of water is 4.18 J/g oC]

Q = 14630 J

Step 4: Energy required to heat water(l) at 100 oC to water(g) at 100 oC

Q = m x delta Hv [Since, this is a phase conversion from liquid to solid, we use heat of vaporisation (Hv)]

Q = 35.0 x 2260 [ Heat of vaporisation of water is 2260 J/g]

Q = 79100 J

Step 5: Energy required to heat water(g) at 100 oC to water(g) at 131.0 oC

Q = mc(delta T)

Q = 35.0x 2.02 x (131 - 100)   [Specific heat of water is 2.02 J/g oC]

Q = 2191.7 J

Now, Total heat = Sum of heats in all the steps

Total heat Q = 1547.7 J + 11690 J + 14630 J + 79100 J + 2191.7 J

Q = 109159.4 J

or

Q = 109.1594 kJ