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Calculating Water's Final Temp after a 345 kJ Heat Boost
Chapter 3, Problem 82P(choose chapter or problem)
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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