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 26.0 g of this metal initially at 20.0 °C?

Answer 1

Answer – We are given, specific heat of metal ,Cm = 0.128 J/(g·°C)

Heat, q = 305 J, mass = 26.0 g , ti = 20.0 °C, tf = ?

We know formula for calculating the heat from specific heat and change in temperature-

q = m *C*∆t

305 J = 26.0 g * 0.128 J/(g·°C) * ( tf – 20.0)^oC

305 J = 3.328 J/ ^oC * ( tf – 20.0)^oC

305 J = 3.328 tf - 66.56

So, 3.328 tf = 305+66.56

                       = 371.56

So, tf = 371.56 / 3.328

          = 112 ^oC

Final temperature if 305 J of heat is added to 26.0 g of this metal initially at 20.0 °C is 112^oC