What is the enthalpy change when 17.0 g of water is cooled from 36.0°C to 6.40°C?
q = delta H = mass x specific heat x (Tfinal-Tinitial).
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To find the enthalpy change when cooling water, you need to use the equation:
ΔH = m × C × ΔT
Where:
ΔH is the enthalpy change
m is the mass of the substance (in this case, water)
C is the specific heat capacity of water
ΔT is the change in temperature
First, let's find the mass of water (m). We are given that 17.0 g of water is cooled.
Next, we need to find the specific heat capacity of water (C). The specific heat capacity of water is typically 4.18 J/g°C.
Finally, let's find the change in temperature (ΔT). The initial temperature of the water is 36.0°C, and it is cooled to a final temperature of 6.40°C.
Now, with all the values determined, we can substitute them into the equation:
ΔH = 17.0 g × 4.18 J/g°C × (6.40°C - 36.0°C)
ΔH = 17.0 g × 4.18 J/g°C × (-29.6°C)
Simplifying the calculation:
ΔH = -1740.32 J
Therefore, the enthalpy change when 17.0 g of water is cooled from 36.0°C to 6.40°C is -1740.32 J.