Question

6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)​

Answer 1

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

Calorimetry

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

Mass of water required

In this case, you know:

• Heat= 92.048 kJ
• Mass of water = ?
• Initial temperature of water= 34 ºC
• Final temperature of water= 100 ºC
• Specific heat of water = 4.186 $\frac{J}{gC}$

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 $\frac{J}{gC}$× m× (100 °C -34 °C)

92.048 kJ = 4.186 $\frac{J}{gC}$× m× 66 °C

m= 92.048 kJ ÷ (4.186 $\frac{J}{gC}$× 66 °C)

m= 0.333 grams

Moles of water required

Being the molar mass of water 18 $\frac{g}{mole}$, that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:

$amount of moles=0.333 gramsx\frac{1 mole}{18 grams}$

amount of moles= 0.0185 moles

Finally, 0.0185 moles of water are required.

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