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how many joules of heat are absorbed when 1.00L of water is heated from 18 degrees C to 85 degrees C?

User Shiko
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Answer: To calculate the amount of heat absorbed by the water, you need to use the formula:

Q = mcΔT

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

Assuming that the density of water is 1 g/mL and its specific heat capacity is 4.184 J/(g·K), the mass of 1.00 L of water is:

m = (1.00 L)(1000 mL/L)(1 g/mL) = 1000 g

The change in temperature is:

ΔT = 85°C - 18°C = 67°C

Now we can plug in these values to find the amount of heat absorbed by the water:

Q = mcΔT

Q = (1000 g)(4.184 J/g·K)(67°C)

Q = 281096 J

Therefore, when 1.00 L of water is heated from 18°C to 85°C, it absorbs approximately 281096 J of heat.

Step-by-step explanation:

User Sanooj
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