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 64.6 g of this metal initially at 20.0 °c?

Answer 1

When we add a certain amount of heat Q to a substance, the temperature of the substance increases by a
`\Delta T` given by

`Q=m C_s \Delta T`
where m is the mass of the substance and Cs is the specific heat capacity of the substance.

By re-arranging the formula, we find

`\Delta T = (Q)/(m C_s)= (305.0 J)/((64.6 g)(0.128 J/gC))=36.9 ^(\circ)C`

So, since the initial temperature of the metal is Ti=20 C, the final temperature is

`T_f = T_i + 36.9 ^(\circ) C=20.0^(\circ) + 36.9^(\circ) C=56.9^(\circ)C`