Question

A nursing mother prepared her baby's milk mixture at 85°C, in a feeding bottle in order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity of 90jk containing 500g of water at 26°C. If the mass of the mixture is 300g calculate the specific heat capacity of the mixture (neglect heat losses and heat capacity of the bottle, specific heat capacity of water=4200jkgk).

Answer 1

First, let's calculate the initial heat content of the milk mixture before it is cooled:

Q1 = mcΔT
Q1 = 300g * c * (85°C - 40°C) [since the initial temperature of the mixture is 85°C]
Q1 = 45c * 300
Q1 = 13500c J

Next, let's calculate the heat lost by the mixture as it is cooled down to 40°C:

Q2 = mcΔT
Q2 = 300g * c * (40°C - 26°C) [since the final temperature of the mixture is 40°C]
Q2 = 14c * 300
Q2 = 4200c J

Since the heat lost by the milk mixture must be equal to the heat gained by the water in the bowl, we have:

Q1 = Q2
13500c = 4200c
13500 = 4200
c = 4200 / 13500
c ≈ 0.311 J/g°C

Therefore, the specific heat capacity of the mixture is approximately 0.311 J/g°C.