Question

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.

Answer 1

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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