A student is preparing to take a bath when she realizes the hot water tap in the bathroom is not working the student go to the kitchen and prepares 10 of 100°C water to mix with 10 L of 10°C water

The final temperature of the bath water will be 55°C

Explanation:

A student is preparing to take a bath when she realizes that the hot water tap is not working.

The student goes to the kitchen and prepares 10 L of 100 degree Celsius water to mix with 10 L of 10 degree Celsius water. What will be the final temperature of the bath water?

Using the relation, Heat lost  = Heat gained (assuming no heat is lost to the surroundings).

Heat lost by hot water = heat gained by cold water

Using the formula of heat, Q = mcΔT

where m = mass of water, c = specific heat capacity of water, ΔT = temperature difference

mass of water = density * volume (density of water = 1 kg/L)

mass of hot water, m₁ = 1 kg/L * 10 L = 10 kg;

mass of cold water, m₂ = 1 kg/L * 10 L = 10 kg;

specific heat capacity of water, c = 4200 j/kg

Let the final temperature of the mixture be T

-m₁cΔT₁ = m₂cΔT₂ ( since heat is lost by the hot water)

-m₁ΔT₁ = m₂ΔT₂

-10 * (T- 100) = 10 * (T - 10)

-10T + 1000 = 10T - 100

20T = 1100

T = 1100/20

T = 55°C

Therefore, the final temperature of the bath water will be 55°C

this would be about 43%

amplitude

hope this