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Maru [420]
3 years ago
10

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 temperatu

re of 105°C
Chemistry
1 answer:
Elden [556K]3 years ago
7 0

Answer:

The total energy required is 121.56 kJ

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\times2.108\times(0-(-10))=40\times2.108\times10=843.2J

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

Energy\:required=mL=40\times336=13440J

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

Energy\:required=ms_{water}\Delta T=40\times4.187\times(100-0)=40\times4.187\times100=16748J

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

Energy\:required=ml=40\times2260=90400J\\

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

Energy\:required=ms_{steam}\Delta T=40\times1.996\times(105-100)=399.2J

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

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Answer:

a. 0,035M

Explanation:

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Thus, solubility is

<em>a. 0,035M</em>

<em></em>

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