How much energy is required to heat a 40 g sample of ice with an initial temperature of -10°C until it reaches a final temperature of 105°C

Question

asked Feb 12, 2020 81.3k views

1 Answer

Mustpax · Answer 1 · 2020-02-17T22:43:57+0000

Answer:

The total energy required is 121.56 kJ

Step-by-step explanation:

The energy required to heat a m = 40g sample of ice is calculated using the following information:

The specific heat of ice =
s_(ice) = 2.108 J/g-K

The specific heat of water =
s_(water) = 4.187 J/g-K

The specific heat of steam =
s_(steam) = 1.996 J/g-K

The latent heat of fusion for ice = L = 336 kJ/kg

The latent heat of vaporisation = l = 2260 kJ/kg

The latent heat of fusion is the amount of energy required to change one kg of ice into water without a change in temperature.

Firstly, we find the energy required to change the temperature of 40 g of ice from -10°C to 0°C.

$Energy\:required=ms_(ice)\Delta T\\ \Delta T\:is\:the\:change\:in\:temperature\\Energy\:required=40*2.108*(0-(-10))=40*2.108*10=843.2J$

Then, the energy required to convert 40g of ice at 0°C to water at 0°C.

$Energy\:required=mL=40*336=13440J$

Then, the energy required to convert water at 0°C to water at 100°C.

$Energy\:required=ms_(water)\Delta T=40*4.187*(100-0)=40*4.187*100=16748J$

The energy required to convert water at 100°C to steam at 100°C.

$Energy\:required=ml=40*2260=90400J\\$

The energy required to convert steam at 100°C to steam at 105°C.

$Energy\:required=ms_(steam)\Delta T=40*1.996*(105-100)=399.2J$

Total Energy required = 843.2 + 13440 + 16748 + 90400 + 399.2 = 121560.4J = 121.56 kJ