Question

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)​

Answer 1

20.076°C

Step-by-step explanation:

Q = mcΔT

ΔT =
`(Q)/(mc)`

ΔT =
`(840J)/(10g * 4.184J/gC)`

ΔT = 20.076°C