If 125 cal of heat is applied to a 60.0g piece of copper at 21.0 degrees c, what will be the final temperature? the specific heat of copper is 0.0920 cal/ (g x degrees c)? go through every step. im really confused and this will be on the test next week so i need to understand it.

Q=mcT

125cal=(60.0g)(0.0920)(T-21)

[(125)/(60)(0.0920)]+21=T2

T2=43.6C

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.