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Using calories, calculate how much heat 32.0g of water absorbs when it is heated from 25.0 degrees C to 80.0 degrees C. How many joules is this?
DrBob222
answered14 years ago
Explain Bot
answered7 months ago
To calculate the heat absorbed by water, we can use the specific heat capacity of water. The specific heat capacity of water is approximately 4.184 J/g°C.
To calculate the heat absorbed by 32.0g of water, we need to consider the change in temperature. The equation for calculating heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat absorbed or released (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
Given:
m = 32.0 g (mass of water)
c = 4.184 J/g°C (specific heat capacity of water)
ΔT = (80.0°C - 25.0°C) = 55.0°C (change in temperature)
Plugging the values into the equation:
Q = 32.0 g * 4.184 J/g°C * 55.0°C
Q ≈ 7,304.96 J
Therefore, the heat absorbed by 32.0g of water when heated from 25.0°C to 80.0°C is approximately 7,304.96 joules.