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

1 Answer
May 11, 2017

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