A piece of metal weighing 52.3 g with a specific heat of 0.54 J9°C-l at 1103.2 °C is dunked into a beaker containing 346.6 mL of water at 25 °C. Assuming all heat is absorbed by the water, what will be the final temperature of the water expressed in °C? Cwater = 4.184 Jg1°C-1 Pwater = 1g/mL Answer in decimal form with 1 decimal place Answer:

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A piece of metal weighing 52.3 g with a specific heat of 0.54 Jg−1∘C−1 at 1103.2 ∘C is dunked into a beaker containing 346.6 mL of water at 25 ∘C. Assuming all heat is absorbed by the water, what will be the final temperature of the water expressed in ∘C? cwater=4.184Jg−1∘C−1 ρwater=1g/mL Answer in decimal form with 1 decimal place
A piece of metal weighing 52.3 g with a specific heat of 0.54 J 91°C-lat 1103.2 °C is dunked into a beaker containing 346.6 mL of water at 25 °C. Assuming all
heat is absorbed by the water, what will be the final temperature of the water expressed in °C?
Cwater = 4.184 Jq1°C-1
Puater= 1g/mL
Answer in decimal form with 1 decimal place
Answer:
Transcribed Image Text:A piece of metal weighing 52.3 g with a specific heat of 0.54 J 91°C-lat 1103.2 °C is dunked into a beaker containing 346.6 mL of water at 25 °C. Assuming all heat is absorbed by the water, what will be the final temperature of the water expressed in °C? Cwater = 4.184 Jq1°C-1 Puater= 1g/mL Answer in decimal form with 1 decimal place Answer:
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