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What is the heat for a reaction of water that masses at 3.3 x 10^4 g that experiences a drop in temperature from 94°C to 64°C?

User Rutrus
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Final answer:

The total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at -30.0 °C is 44,801.44 J. can be calculated using the equation: q = mCΔT where: q is the total heat released m is the mass of water C is the specific heat of water ΔT is the change in temperature.

Step-by-step explanation:

The total amount of heat released can be calculated using the equation: q = mCΔT where: q is the total heat released m is the mass of water C is the specific heat of water ΔT is the change in temperature Using the given values: Mass of water (m): 94.0 g Change in temperature (ΔT): 80.0 ℃ - (-30.0 ℃) = 110.0 ℃ Specific heat of water (C): 4.184 J/g ℃ Substituting these values into the equation: q = 94.0 g x 4.184 J/g ℃ x 110.0 ℃, q = 44,801.44 J the total amount of heat released when 94.0 g water at 80.0 ℃ cools to form ice at -30.0 ℃ is 44,801.44 J.

The total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at -30.0 °C is 44,801.44 J. can be calculated using the equation: q = mCΔT where: q is the total heat released m is the mass of water C is the specific heat of water ΔT is the change in temperature Using the given values: Mass of water (m): 94.0 g Change in temperature (ΔT): 80.0 ℃ .

User Suyash
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