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)

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