An object with a mass of #160 g# is dropped into #880 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #18 ^@C#, what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-02-18T09:27:00

The specific heat is #= 3.453 KJkg^-1°C^-1#

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water, # Delta T_w=18ºC#

For the object #DeltaT_o=120ºC#

# M_o C_o (DeltaT_o) = Mw C_w (Delta_w)#

#C_w=4.186KJkg^-1ºC^-1#

#0,160 C_o120 = 0,880* 4.186 *18#

#C_o = (0.880*4.186*18)/(0.160*120) = 3.453 KJkg^-1°C^-1#