Question

The specific heat of a certain type of metal is 0.128J/g*C What is the final temperature if 305 J of heat is added to 93.5g of this metal, initially at 20.0 C.

Tfinal = ___________C

Answer 1

Using. Heat, q = mC(Tf - Ti)

Where, q = 305 J, m = 93.5 g, C = 0.128 J/g°C, Ti = 20°C

So, 305 = 93.5×0.128(Tf - 20)

Tf - 20 = 305/93.5×0.128

Tf - 20 = 25.5

Tf = 25.5+20

= 45.5° C