Solving the Hardest Thermal Process Question: Find W/Q Ratio

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The discussion centers on a challenging thermodynamics problem involving the W/Q ratio for water heated in an open pan at 1 atm pressure. The user struggles to determine the correct value for Q, initially suggesting Q = cm delta T but doubts its accuracy. Clarification is sought on whether the conversion from calories to Joules was performed correctly. The conversation highlights the importance of accurately calculating both work done (W) and heat absorbed (Q) in thermodynamic processes. Assistance is requested to resolve the confusion surrounding the calculations.
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This question is considered one of the hardest in this chapter regarding thermodynamics. I spent countless time trying to figure it out but i was never close to the answer. The question is:

Water is heated in an open pan where the air pressure is 1atm. WAter remains a liquid but expanded by a small amount when it was heated. FIND THE RATIO OF WORK DONE BY WATER TO THE HEAD ABSOBED BY THE WATER.

So that means:

W / Q

where W = P(Vf - Vi)
but what's Q? I tried Q = cm delta T but i was wrong. if anyone can help me i'd appreciate it thanks.
 
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Q=cm delta T seems right. Did you convert calories to Joules?
 

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