When 155 g of water at 20 ᵒC is mixed with 75 g of water at 80 ᵒC, assuming no heat is lost to the surrounding what is the final temperature of the water. (Specific heat capacity

of water is 4.18 J/°C·g)​

39.57 °C

Explanation:

Applying,

Heat gain = Heat lost.

cm'(t₃-t₁) = cm(t₂-t₃) Equation 1

Where c = specific heat capacity of water, m = mass of the hotter water, m' = mass of the colder water, t₁ = initial temperature of the colder water, t₂ = initial temperature of the hotter water, t₃ = Final Temperature.

Equation 1 above can futher be simplified to

m'(t₃-t₁) = m(t₂-t₃) Equation 2

From the question,

Given: m' = 155 g, m' = 75 g, t₁ = 20 °C, t₂ = 80 °C

Substitute these values into equation 2

155(t₃-20) = 75(80-t₃)

Solve for t₃

155t₃- 3100 = 6000-75t₃

155t₃+75t₃ = 6000+3100

230t₃ = 9100

t₃ = 9100/230

t₃ = 39.57 °C

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.