# What is the final temperature if a metal at #75^@ "C"# is dropped into #"185 g"# of water starting at #25^@ "C"#? The heat capacity of the metal is #"55 J/g"*""^@ "C"#.

##### 2 Answers

This is one of those questions where solving it in general would help you more with not messing up on the math.

#T_(f) = [m_wC_wT_(iw) + c_mT_(im)]/(c_m + m_wC_w)#

I get

You know that there is conservation of energy, so

#q_m + q_w = 0# ,

#q_m = -q_w#

where

#q_m = m_mC_mDeltaT_m#

#q_w = m_wC_wDeltaT_w# where

#m# is mass in#"g"# ,#C# is specific heat capacity in#"J/g"^@ "C"# or#"J/g"cdot"K"# , and#DeltaT# is the change in temperature in either#""^@"C"# or#"K"# .

#m_mC_mDeltaT_m = -m_wC_wDeltaT_w#

The metal is hotter, so dropping it into water must cool it down. So,

#m_mC_m(T_f - T_(im)) = -m_wC_w(T_f - T_(iw))#

#m_mC_m(T_f - T_(im)) = m_wC_w(T_(iw) - T_f)#

Distribute the terms:

#m_mC_mT_f - m_mC_mT_(im) = m_wC_wT_(iw) - m_wC_wT_f#

Now get the

#m_mC_mT_f + m_wC_wT_f = m_wC_wT_(iw) + m_mC_mT_(im)#

Finally, factor out

#T_(f)(m_mC_m + m_wC_w) = m_wC_wT_(iw) + m_mC_mT_(im)#

#T_(f) = [m_wC_wT_(iw) + m_mC_mT_(im)]/(m_mC_m + m_wC_w)#

#color(blue)(T_(f) = [m_wC_wT_(iw) + c_mT_(im)]/(c_m + m_wC_w))# where

#m_mC_m = c_m# , the heat capacity in#"J/"^@ "C"# .

So now, we can plug the numbers in and solve for the final temperature.

Let the final temperature be

Heat lost by metal

Heat gained by water

So by calorimetric principle

Now dividing both sides by 5 we get