Question

When 168 joules of heat is added 4 grams of water at 283 K, what is the resulting temperature?

Answer 1

`293 K`

Explanation:

The specific heat formula:
`Q=c*m*Delta T`, where `Q` is the amount of heat transferred, `c` is the specific heat capacity of the substance, `m` is the mass of the object, and `Delta T` is the change in temperature. In order to solve for the change in temperature, use the formula

`Delta T=Q/(c_(water)*m)`

The standard heat capacity of water, `c_(water)` is `4.18* J*g^(-1)*K^(-1)`.

And we get `Delta T=(168*J)/(4.18 *J*g^(-1)*K^(-1)*4*g)=10.0 K`

Since `Q>0`, the resulting temperature will be `T_(f)=T_(i)+Delta T=283 K+10.0K=293K`
(pay special attention to significant figures)

Additional resources on Heat Capacity and Specific Heat:
https://www.ck12.org/chemistry/Heat-Capacity-and-Specific-Heat/lesson/Heat-Capacity-and-Specific-Heat-CHEM/?referrer=concept_details