What is the final temperature in degrees centigrade of 100g of water at 30 c if it is mixed with 50 g of water at 0?

The final temperature = 293 K or 20 °C

Explanation:

Heat gained by water at 0°C = heat lost by water at 30°C

c₁m₁ (T₃ - T₁) = c₂m₂ (T₂-T₃) ... Equation 1

Making T₃ the subject of the equation,

T₃ = (c₂m₂T₂ + c₁m₁T₁) / (c₁m₁+c₂m₂) ... Equation 2

Where m₁ = mass of water at 0°C, c₁ = specific heat capacity of water at 0°C, c₂ = specific heat capacity of water at 30°C, m₂ = mass of water at 30°C, T₁ = initial temperature of water at 0°C, T₂ = initial temperature of water at 30°C, T₃ = final temperature.

Given: m₁ = 50 g = (50/1000) kg = 0.05 kg, m₂ = 100 g = (100/1000) kg = 0.1 kg., T₁ = 0°C = 273 K, T₂ = 30°C = (30+273) = 203 K

Constants: c₁ = c₂ = 4200 J/kg. K

Substituting these values into equation 2,

T₃ = (4200*0.1*303 + 4200*0.05*273) / (4200*0.1 + 4200*0.05)

T₃ = (127260 + 57330) / (420+210)

T₃ = 184590/630

T₃ = 293 K.

Therefore the final temperature = 293 K or 20 °C