A glass of cold water with a mass of 200 grams at a temperature of 20 Celsius mixed with 100 grams of hot water at 80 Celsius. If the glass is deemed not to receive heat, what is the temperature of the mixture of hot water and cold water? (Heat water type + 1 cal / g Celsius)
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Since both of the same type of liquid liquids are the same:
Tc = (m₁T₁ + m₂T₂) / (m₁ + m₂)
= (0.1 kg) (80 ° C) + (0.2 kg) (20 ° C)] / (0.1 kg + 0.2 kg) = 40 ° C
Since both of the same type of liquid liquids are the same:
Tc = (m₁T₁ + m₂T₂) / (m₁ + m₂)
= (0.1 kg) (80 ° C) + (0.2 kg) (20 ° C)] / (0.1 kg + 0.2 kg) = 40 ° C
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