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If you mix 100 milliliters of 10 degrees Celsius of water and 100 milliliters of 90 degrees celsius of water what will be your final temperature of the mixture

User Miojamo
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2 Answers

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Answer:

Final temperature of the mixture is 50°C

Step-by-step explanation:

This is a typical calorimetry problem, where the heat yielded by a compound is immediately won by another.

Water at 90°C raise the temperature of water at 10°

Density of water = 1g/mL

Mass of water = 100 g, because the volume is 100mL

If the two pieces of water are in a calorimeter where heat is not released or gained, the sum of both heats is 0

4.18 J/g°C . 100 g ( T°final - 90°C) + 4.18 J/g°C . 100 g ( T°final - 10°C) = 0

418 J/°C T°final - 37620 J + 418 J/°C T°final - 4180J = 0

836 J/°C T°final - 37620J - 4180J = 0

836 J/°C T°final = 41800J

T°final = 41800J / 836 °C/J = 50°C

User Jedivader
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7.3k points
0 votes

Answer:

Final temperature of the mixture is 50°C

Step-by-step explanation:

This is a typical calorimetry problem, where the heat yielded by a compound is immediately won by another.

Water at 90°C raise the temperature of water at 10°

Density of water = 1g/mL

Mass of water = 100 g, because the volume is 100mL

If the two pieces of water are in a calorimeter where heat is not released or gained, the sum of both heats is 0

4.18 J/g°C . 100 g ( T°final - 90°C) + 4.18 J/g°C . 100 g ( T°final - 10°C) = 0

418 J/°C T°final - 37620 J + 418 J/°C T°final - 4180J = 0

836 J/°C T°final - 37620J - 4180J = 0

836 J/°C T°final = 41800J

T°final = 41800J / 836 °C/J = 50°C

User Hjvcghcghchg
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