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3. What is the change in temperature after 840 Joules is absorbed by 10.0g of water is heated?
(Specific heat of water is 4.184 J/g °C)​

Respuesta :

Answer:

20.076°C

Explanation:

Q = mcΔT

ΔT = [tex]\frac{Q}{mc}[/tex]

ΔT = [tex]\frac{840J}{10g * 4.184J/gC}[/tex]

ΔT = 20.076°C

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