An object with a mass of #24 g# is dropped into #6400 mL# of water at #0^@C#. If the object cools by #60 ^@C# and the water warms by #16 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
G_Ozdilek
Nov 3, 2016

71.11 kcal/(kg*degrees C)

Explanation:

#m*c*Delta(t)# equation can be used. For water #6.4*16*1# will be equal to #0.024*60*C#. When you solve this, you will get #C = 71.11 (kcal)/(kg*degrees C)#

Related questions
See all questions in Specific Heat
Impact of this question
1223 views around the world
You can reuse this answer
Creative Commons License