Question

Approximately 20.0gm of milk at 6.0oC is added into a cup containing 270.0 gm of weak tea. The specific heat of weak tea is 3.91 x 103J kg-1 oC-1 and the final temperature of the milk - tea mixture is 85.0oC. Given the initial temperature of the weak tea is 90.0oC, what is the specific heat of milk?

Answer 1

4161 J/kg·°C

Step-by-step explanation:

We can use the principle of conservation of energy to solve this problem, which states that the total heat energy in a closed system is constant. The heat lost by the tea is equal to the heat gained by the milk.

Let's first calculate the heat lost by the tea:

Q(tea) = mcΔT

Q(tea) = (0.27 kg)(3910 J/kg·°C)(90.0°C - 85.0°C)

Q(tea) = 6555 J

where m is the mass of tea, c is the specific heat of tea, and ΔT is the change in temperature.

Next, let's calculate the heat gained by the milk:

Q(milk) = mcΔT

Q(milk) = (0.02 kg)(c)(85.0°C - 6.0°C)

Now we can equate the two expressions:

Q(tea) = Q(milk)

6555 J = (0.02 kg)(c)(79.0°C)

Solving for c, we get:

c = 4161 J/kg·°C

Therefore, the specific heat of milk is approximately 4161 J/kg·°C.