Question: What is the final temp of 25.0 grams of water with initial temp of 22.0 degrees C after it absorbs 454 J of heat?

What is the final temp of 25.0 grams of water with initial temp of 22.0 degrees C after it absorbs 454 J of heat

 

Answer:

The final temperature of the water can be calculated using the equation:

 

ΔT = Q / (mCp)

 

Where ΔT is the change in temperature, Q is the heat absorbed (in joules), m is the mass of water (in kilograms), and Cp is the specific heat capacity of water (in joules per kilogram per degree Celsius). The specific heat capacity of water is approximately 4.18 J/g°C.

 

First, we need to convert the mass of water to kilograms:

 

m = 25.0 g / 1000 g/kg = 0.025 kg

 

Next, we can calculate the change in temperature:

 

ΔT = Q / (mCp) = 454 J / (0.025 kg * 4.18 J/g°C) = 27.7°C

 

Finally, we can add the change in temperature to the initial temperature to find the final temperature:

 

Tfinal = 22.0°C + ΔT = 22.0°C + 27.7°C = 49.7°C

 

Therefore, the final temperature of the 25.0 grams of water after it absorbs 454 J of heat is 49.7°C.

 

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