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

1 Answer
May 11, 2017

Answer:

#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