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How many joules of heat is released when 29 grams of water is cooled from 293k to 283

User Zeyad
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1 Answer

4 votes

Final answer:

The heat released is -438,692 Joules.

Step-by-step explanation:

To calculate the total amount of heat released when water cools and forms ice, we need to use the equation:

Q = mcΔT

where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Given that the mass of water is 94.0 g, the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is (-30.0) - 80.0 = -110.0°C, we can substitute these values into the equation:

Q = (94.0 g)(4.18 J/g°C)(-110.0°C)

Simplifying the equation, we get:

Q = -438,692 J

Therefore, the total amount of heat released when 94.0 g water cools to form ice at -30.0 °C is -438,692 Joules.

User Matt Andrzejczuk
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