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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)?

please go through every step. im really confused and this will be on the test next week so i need to understand it.

Respuesta :

prosborough198p2ub41

Q=mcT

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

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

T2=43.6C

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