When you take a bath, how many kilograms of hot water (40.8 °C) must you mix with cold water (11.3 °C) so that the temperature of the bath is 30.9 °C? The total mass of water (hot plus cold) is 190 kg. Ignore any heat flow between the water and its external surroundings.

Q₁=m₁cΔT₁

Q₂=m₂cΔT₂
m₁cΔT₁=m₂cΔT₂
m₁ΔT₁=m₂ΔT₂
m₁ΔT₁=m₂ΔT₂
m (40.8 -30.9)=(190-m)(30.9-11.3)
Solve for ‘m’

To solve this problem, we need to use the principle of conservation of energy. The amount of heat gained by the cold water and the amount of heat lost by the hot water will be equal when the bathwater reaches the desired temperature.

The heat gained or lost by an object can be calculated using the formula:

Q = mcΔT

Where:
Q = heat gained or lost (in joules)
m = mass of the object (in kilograms)
c = specific heat capacity of the object (in joules per kilogram-degree Celsius)
ΔT = change in temperature (in degrees Celsius)

In this case, we want to find the mass of hot water (m_hot) that needs to be mixed with cold water. Let's assume the specific heat capacity of water is the same for both hot and cold water.

We can split the problem into two parts: the heat gained by the cold water and the heat lost by the hot water.

For the cold water:

Q_cold = m_cold * c * ΔT

For the hot water:

Q_hot = m_hot * c * ΔT

Since the total mass of water is given as 190 kg, we can calculate the mass of cold water:

m_cold = 190 kg - m_hot

Now, we set up an equation to equate the heat gained by the cold water with the heat lost by the hot water:

m_cold * c * ΔT = m_hot * c * ΔT

Canceling out the specific heat capacity and change in temperature:

m_cold = m_hot

Substituting the value of m_cold with (190 kg - m_hot):

190 kg - m_hot = m_hot

Rearranging the equation:

2m_hot = 190 kg

Solving for m_hot:

m_hot = 190 kg / 2

m_hot = 95 kg

Therefore, you would need to mix 95 kilograms of hot water with the cold water in order to attain a bath temperature of 30.9 °C.