Question

CP10 15-18

1. joules to condense 127 g of steam at 100 ∘C and to cool the liquid to 18.0 ∘C

2. joules needed to melt a 529-g ice sculpture at 0 ∘C and to warm the liquid to 10.5

3 .kilojoules released when 76.5 g of steam condenses at 100 ∘C, the liquid cools to 0 ∘C, and freezes

4. joules to warm 40.5 mL of water (density=1.00g/mL) from 23.5 ∘C to 100 ∘C and vaporize it at 100 ∘C

∘C

Express your answer with the appropriate units.

Answer 1

1. Specific latent heat of water = 2260 j/g. = L

Mass = m = 127 g.

Specific heat capacity of water = 4.18 j/g℃ = c.

Temperature change =(18-0)℃

So total heat required= mc(18-0) + mL

= 127*4.18*18 + 127*2260

=296575 j.

2. Mass of sculpture= m = 529 g.

Specific heat capacity of ic = 2.1 08 j/g℃.

Specific latent heat of condensation = 336 j/g.

Temperature change =10.5℃

Total heat required = mc*10.5 + mL

=529*2.108*10.5+529*336

=189453 j.

3. Mass = m = 76.5 g.

L1 = 2260 j/g(steam to liquid water)

L2 = 336 j/g(liquid water to ice)  

So heat released = mL1 +mc*100 + mL2

= 76.5( 2260 + 418 + 336)

= 230571 j.( Heat released)

4. Mass of water = 40.5 g

Temperature change = 100 - 23.5 = 76.5 ℃.

c =4.18 j/g℃.

L = 2260 j/g.

Heat required = mc*76.5 + mL = 40.5(2260 + 76.5*4.18)

= 40.5 * 2580

= 104490 j.