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How much water can be heated 20.0 °C with 1,000.0 J of heat energy?​

User Jackberry
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Answer:

The amount of water that can be heated by 1,000.0 J of heat energy depends on the mass of water and the specific heat capacity of water.

Assuming the water is at an initial temperature of 20.0°C, we can use the formula:

Q = mcΔT

Where:

Q = heat energy (Joules)

m = mass of water (in grams)

c = specific heat capacity of water (4.184 J/g°C)

ΔT = change in temperature (final temperature - initial temperature)

Rearranging the formula to solve for the mass of water:

m = Q / (c*ΔT)

Plugging in the given values:

m = 1000 J / (4.184 J/g°C * (final temperature - 20.0°C))

Assuming the final temperature is 100.0°C (the boiling point of water at standard pressure), the calculation becomes:

m = 1000 J / (4.184 J/g°C * (100.0°C - 20.0°C))

m = 1000 J / (4.184 J/g°C * 80.0°C)

m = 2.39 grams

Therefore, 1,000.0 J of heat energy can heat 2.39 grams of water from 20.0°C to 100.0°C.

User Schmalzy
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