A coffee cup calorimeter contains 155.47 g of water at 22.50 deg C. A 65.676 g...

  

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A coffee cup calorimeter contains 155.47 g of water at 22.50 \deg C. A 65.676 g piece of iron is heated to 101.13 \deg C. The piece of iron is added to the coffee cup caloriemter and the contentsReach thermal equilibrium at 25.68 C. thespecific heat capacity of water is 4.184 - 2224.91J heat is lost by the piece of iron 2069J heat is gained by the water The difference between the heat lost by the piece of iron and the heat gained by the water is due to heat transfer to the styrofoan and the heat required to raise the temperature of the calorimeter. What is the heat capacity of the styrofoam calorimeter in joule per kelvin (R) ? And What would be the final temperature of the system if all of the heat lost by the iron was absorbed by the water? A coffee cup calorimeter contains 155.47 g of water at 22.50 \deg C. A 65.676 g piece of iron is heated to 101.13 \deg C. The piece of iron is added to the coffee cup caloriemter and the contentsReach thermal equilibrium at 25.68 C. thespecific heat capacity of water is 4.184 - 2224.91J heat is lost by the piece of iron 2069J heat is gained by the water The difference between the heat lost by the piece of iron and the heat gained by the water is due to heat transfer to the styrofoan and the heat required to raise the temperature of the calorimeter. What is the heat capacity of the styrofoam calorimeter in joule per kelvin (R) ? And What would be the final temperature of the system if all of the heat lost by the iron was absorbed by the water?

Answer rating: 100% (QA)

First lets calculate the heat transferred to the styrofoam calorimeter We know that the heat lost by ... View the full answer