In: Chemistry
If you combine 360.0 mL of water at 25.00 °C and 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
Let final temperature of mixture is T oC.
As density of water is 1.00 g/mL, Mass of water in gram is numerically equal to volume in mL.
Let for cooler water sample,
Mass = m = 360.0 g
Initital temperature = 25.00 oC
Final temperature = T oC
Increase in temperatue = (T-25.00) oC
Heat gain by water = Mass * Sp. heat of water * Increase in temperature
= 360.0 g * Sp.Heat of water * (T-25.00) oC
Then for hotter water sample,
Mass = m = 100.0 g
Initital temperature = 95.00 oC
Final temperature = T oC
Decrease in temperatue = (95.00-T) oC
Heat loss by water = Mass * Sp. heat of water * Decrease in temperature
= 100.0 g * Sp.Heat of water * (95.00-T) oC
On mixing two smpple we have,
Heat gain by water = Heat loss by water
360.0 g * Sp.Heat of water * (T-25.00) oC = 100.0 g * Sp.Heat of water * (95.00-T) oC
360.0 * (T-25.00) = 100.0 * (95.00-T)
360.0 T- 360.0 * 25.00 = 100.0*95.00 - 100.0T
460.0 T = 100.0*95.00 + 360.0 * 25.00
460.0 T = 18500
T = 40.22 oC
Final temperature of mixture is 40.22 oC.
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