The following physical constants are for water, H2O. The specific heat capacity of the solid = 2.09 J/g oC
The specific heat capacity of the liquid = 4.18 J/g oC
The specific heat capacity of the vapor = 2.09 J/g oC
∆Hfus = 6.02 kJ/mol; ∆Hvap = 40.7 kJ/mol Freezing point = 0.0oC; Boiling point = 100.0oC
How much heat(in kJ) is required to warm 10.0 grams of ice at -5.0oC to a temperature of 70.0oC?

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

here;

mass, m = 10 g

specific heat capacity of ice,

change in temperature,

Amount of heat required to melt the ice at 0°C:

where,

we know that no. of moles is = (wt. in gram) (molecular mass)

Now, the heat required to bring the water to 70°C from 0°C:

specific heat of water,

change in temperature,

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

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in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.