UMCI JUU 2015 WIL PIUTUU VUA. When 136.0 mL of water at 26.0°C is mixed with 56.0 mL of water at 85.0°C, what i...

Question

Question

Answer 1

Answer #1

m₁ grams of a substance at T₁ temperature is mixed with

m₂ grams of a substance at T₂ temperature, final temperature of the mixture is

T = mi71 + m2 12 mi + m2

Given that

Temperature of the water (T₁) : 26°C = 26+273 = 299 K

Volume of the water : 136 mL

Since density of water is 1 g/mL, mass of the water is 136 grams

Volume of 2nd water sample = 56 mL

Since density of water is given 1 g/mL, mass of 2nd water sample is 56 grams

Temperature of the 2nd water sample = 85°C

= 85+273 = 355 K

T =- 1369 * 299K + 56g * 355K 1369 +569

T = 315.3 K = 315.3 - 273 = 42.3°C

Answer 2

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