Question

Enter your answer in the provided box. When 131.0 mL of water at 26.0°C is mixed with 41.0 mL of water at 85.0°C, what is the final temperature? (Assume that no heat is lost to the surroundings; d of water is 1.00 g/mL.)

Answer 1

Here is the Solution:

The relationship betweet heat and temperature change is as follows. Where, q = is the heat q = miciat m = mass of water. C = Specific heat of water 4.184 79.k at = change in temperature (T2-To) * The density of water is 1.00 g/mL . This means, the volume of water is equivalent to the mass of water, · Mass of water at higher temperature (85.0°C) = 41 g · Mass of water at lower temperature (26.0°C) = 131 g The heat released by the water at higher temperature is gained by the water at lower temperature, -(c.m.at) mighettore = (com.at semperature -4.184 8fg, K x 428 X (12=85%)] - [ 4. 184 7g. K x 1325 x{12-268] =D -41 (T2-85°C) = 131 (T2 - 26 °C) - 41T2 + 3485 C = 131T2 - 3406 " * 13172 +41 T2 = 3485°C + 3406 °C = 172 T2 = 6891 °C i Tz = 6891°C At higher temperature Lower temperature . A A A 172 Tz = [ 40.06% Hence, 40.06°C is the final temperature,

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