Find the temperature of a mixture obtained by mixing 32 grams of water at a temperature of 22 degrees Celsius with 88 grams of water at a temperature of 75 degrees Celsius.

60.87 °C

Explanation:

Applying,

Heat lost = heat gain

cm'(t₂-t₃) = cm(t₃-t₁) Equation 1

Equation 1 can futher be simplified to

m'(t₂-t₃) = m(t₃-t₁)Equation 2

Where m' = mass of the hot water, m = mass of the cold water, t₁ = initial temperature of the cold water, t₂ = initial temperature of the hot water, t₃ = temperature of the mixture.

From the question,

Given: m' = 88 g, m = 32 g, t₁ = 22°C, t₂ = 75°C

Substitute these values into equation 2

88(75-t₃) = 32(t₃-22)

6600-88t₃ = 32t₃-704

32t₃+88t₃ = 6600+704

120t₃ = 7304

t₃ = 60.87 °C

temp rises, and expansion

no. dot product   =0 when vecs at right angles