How much heat is absorbed when a 10.0-g sample of water is changed from 23.0⁰C to 32.0⁰C?

376.56J

Explanation:

The amount or quantity of heat absorbed can be calculated using the formula:

Q = m × c × ∆T

Where; Q = amount of heat absorbed (J)

m = mass of water (g)

c = specific heat capacity of water (J/g°C)

∆T = change in temperature (°C)

According to the provided information in this question; m = 10.0g, c = 4.184J/g°C, final temperature = 32°C, initial temperature = 23°C

Q = m × c × ∆T

Q = 10 × 4.184 × (32 - 23)

Q = 41.84 × 9

Q = 376.56J

now split the equation in half

CH4+O2                 I                  CO2 + 4H2O

C = 1                      I                  C = 1                    

H = 4                      I                  H = 8

O = 2                      I                 O = 3

CH4+ 2O2 = CO2 + 2H20