An object with a mass of #21 g# is dropped into #400 mL# of water at #0^@C#. If the object cools by #150 ^@C# and the water warms by #14 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
MeneerNask
Feb 19, 2016

The heat energy lost by the object is the same as the energy gained by the water, or #Q_o=Q_w#

Explanation:

#c_o*m_o*DeltaT_o=c_w*m_w*DeltaT_w#

Now fill in what we know (#c_w=4.2# and #1mL=1g# water):
#c_o*21*150=4.2*400*14#

Work this out:
#c_o=(4.2*400*14)/(21*150)~~7.47#

This is a ridiculously high outcome, as no known material has a specific heat this high. Probably a #0# was lost or added in the problem.
With a mass of #210g# or an amount of water of #40mL# the outcome would be #0.747#, which is more plausible.

Related questions
See all questions in Specific Heat
Impact of this question
1267 views around the world
You can reuse this answer
Creative Commons License