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How many kilojoules of heat are required to heat 1.37kg of water from 21.3 to 89.5

Sagot :

q= ? kJ                  
m= 1.37 kg × [tex] \frac{1000 g}{1kg} [/tex] =1370 g
                      
c=4.18 J/g° C
ΔΤ= 89.5-21.3= 68.2˚C
? J = (1370 g) (4.18 J/g˚C) (68.2˚C)
q = 390,554 J× [tex] \frac{1kJ}{1000J} [/tex]
                 
                    q = 391 kJ
Mickymike92idn

The quantity of Heat, Q, required to heat 1.37kg of water from 21.3 to 89.5 is 390.55 kJ.

What is quantity of Heat?

The quantity of Heat is the amount of heat energy added or removed from a substance.

  • Quantity of Heat = mass × specific heat capacity × temperature change

The quantity of Heat, Q, required to heat 1.37kg of water from 21.3 to 89.5 is calculated as follows:

  • Q = mcΔΤ

mass of water = 1.37 kg = 1370 g

specific heat capacity of water = 4.18 J/g° C

Temperature difference, ΔΤ= 89.5-21.3= 68.2˚C

Q = 1370 g × 4.18 J/g˚C × 68.2˚C

Q = 390,554.12

Q = 390.55 kJ

Therefore, the quantity of Heat, Q, required to heat 1.37kg of water from 21.3 to 89.5 is 390.55 kJ.

Learn more about quantity of Heat at: https://brainly.com/question/490326

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