Question

A mass of (200 g) of hot water at (75.0°C) is mixed with cold water of mass M at (5.0°C). The final temperature of the mixture is (25.0°C). What is the mass of the cold water (M)?

Answer 1

The mass of the cold water, given the data from the question is 500 g

Data obtained from the question

• Mass of warm water (Mᵥᵥ) = 200 g
• Temperature warm water (Tᵥᵥ) = 75 °C
• Temperature of cold water (T꜀) = 5 °C
• Equilibrium temperature (Tₑ) = 25 °C
• Specific heat capacity of the water = 4.184 J/gºC
• Mass of cold water (M꜀) =?

How to determine the mass of the cold water

Heat loss = Heat gain

MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – T꜀)

200 × 4.184 (75 – 25) = M꜀ × 4.184(25 – 5)

41840 = M꜀ × 83.68

Divide both side 83.68

M꜀ = 41840 / 83.68

M꜀ = 500 g

Learn more about heat transfer:

brainly.com/question/6363778

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