Question

A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the water? Group of answer choices

Answer 1

The final temperature of the water is 80.12°C.

Step-by-step explanation:

To find the final temperature, we can use the principle of conservation of energy. The heat gained by the colder water is equal to the heat lost by the hotter water. The heat gained or lost can be calculated using the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

In this case, the heat gained by the colder water is equal to the heat lost by the hotter water. We can write the equation as:

mcΔT_hot + mcΔT_cold = 0

Using the given values, we have:

(60 L) (1000 g/L) (30°C - T_f) + (5.0 L) (1000 g/L) (T_f - 80°C) = 0

Simplifying the equation and solving for T_f, we find:

T_f = (60 L) (1000 g/L) (30°C) / [(60 L) (1000 g/L) + (5.0 L) (1000 g/L)] + 80°C

T_f = 0.12°C + 80°C = 80.12°C

Answer 2

33.85°C

Step-by-step explanation:

From the question,

Heat lost by the hotter water = heat gained by the colder water

cm'(t₂-t₃) = cm(t₃-t₁)................. Equation 1

Where c = specific heat capacity of water, m' = mass of hot water, m = mass of cold water, t₁ = Initial temperature of cold water, t₂ = Initial temperature of hot water, t₃ = final temperature of the mixture.

But since the density of water is constant, and mass varies directly as volume, We can replace the mass of water with the volume of water. i.e,

cv'(t₂-t₃) = cv(t₃-t₁)................. Equation 2

Where v' and v are the volume of hot water and cold water respectively

make t₃ the subject of the equation

t₃ = (v't₂+vt₁)/(v'+v)............ Equation 3

Given: v' = 5.0 L, v = 60 L, t₁ = 30°C, t₂ = 80°C

Substitute these values into equation 3

t₃ = (5×80+60×30)/(60+5)

t₃ = 2200/68

t₃ = 33.85°C