Calculate Tf When 77g Ice & 300g Water at 0°C & 50°C Mix

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SUMMARY

The final temperature (Tf) when mixing 77 grams of ice at 0 °C with 300 grams of water at 50 °C can be calculated using the principles of conservation of thermal energy. The specific heat of water is 1.00 cal/(gram*°C), and the heat of fusion for ice is 79.7 cal/gram. The energy lost by the warm water must equal the energy gained by the ice as it melts and warms up. This calculation involves setting the heat lost by the water equal to the heat gained by the ice, leading to a definitive Tf value.

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Determine the final temperature Tf, that results when 77 grams of ice at 0 °C are mixed with 300 grams of liquid water at 50 °C.
Specific heat of water: c = 1.00 cal/(gram*°C).
Heat of fusion for the ice - liquid water transition: cF = 79.7 cal/gram

I have tried this several times and several different ways and cannot solve the answer.
 
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dmolson said:
Determine the final temperature Tf, that results when 77 grams of ice at 0 °C are mixed with 300 grams of liquid water at 50 °C.
Specific heat of water: c = 1.00 cal/(gram*°C).
Heat of fusion for the ice - liquid water transition: cF = 79.7 cal/gram

I have tried this several times and several different ways and cannot solve the answer.

Tell us how you think it should be done, even if you did get the wrong answer. Then we can help you.
 
What are your thoughts on this?

Apply conservation of thermal energy. Take in the before and after situation and equate the increase of wnergy with the loss of energy.
 

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