Question

The specific heat of a certain type of metal is 0.128 J/(g-°C). What is the final temperature if 305 J of heat is added to 79.3 g of

this metal, initially at 20.0 °C?
Tfinal =
°C

Answer 1

final temp is 79.1 ºC

Step-by-step explanation:

mass * specific heat * temperature change

305 J = (40.3 g)(0.128 J/g °C) ΔT.

ΔT = 305/5.158 = 59.1 ºC

20.0 + 59.1 = 79.1 ºC