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Calculating Water's Final Temp after a 345 kJ Heat Boost

Chapter 3, Problem 82P

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QUESTION:

A 45-kg sample of water absorbs 345 kJ of heat. If the water was initially at \(22.1\ ^{\circ}\mathrm{C}\), what is its final temperature?

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QUESTION:

A 45-kg sample of water absorbs 345 kJ of heat. If the water was initially at \(22.1\ ^{\circ}\mathrm{C}\), what is its final temperature?

ANSWER:

 

Step 1 of 2

Given :

Sample of amount of water = 45 kg

Heat absorbed = 345 kJ

Initial temperature of water = 22.1 °C.

Final temperature  = ?

To find the temperature change, we can rearrange the formula \(\mathrm{q}=\mathrm{mc} \Delta \mathrm{T}\) where 'q' is heat energy, 'm' is mass, 'c' is specific heat capacity, and '\(\Delta \mathrm{T}\)' is change in temperature

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Calculating Water's Final Temp after a 345 kJ Heat Boost
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Discover how to determine the final temperature of a water sample after heat absorption. Utilizing the concept of specific heat capacity and the formula q = mc?T, we calculate the change in temperature and reveal its new state. Experience the transformation from an initial tepid state to a drastically altered thermal condition.

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