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How much energy is required to raise the temperature of 10.7 grams of gaseous helium from 22.1 °C to 39.4 °C ?

Sagot :

Neetooidn

Answer:

Q = 2640.96 J

Explanation:

Given data:

Mass of He gas = 10.7 g

Initial temperature = 22.1°C

Final temperature = 39.4°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 39.4°C - 22.1°C

ΔT = 17.3°C

Q = 10.7 g× 14.267 J/g.°C ×  17.3°C

Q = 2640.96 J

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