An object with a mass of #20 g# is dropped into #750 mL# of water at #0^@C#. If the object cools by #90 ^@C# and the water warms by #18 ^@C#, what is the specific heat of the material that the object is made of?

2 Answers
Mar 4, 2018

suppose,the specific heat of the object is #s Cg^-1^@C^-1#

So,during this process of achieving thermal equilibrium,the heat energy released by the object was taken by the water.

So,we can write,

#20*s*90 = 750*1*18# (using, #H=msd theta#,where, #d theta# is the change in temperature,and specific heat for water is #1# CGS unit and weight of #750mL# of water is #750g#)

So,we get, #s=7.5 Cg^-1^@C^-1#

Mar 4, 2018

Answer:

The specific heat is #=31.4 kJkg^-1K^-1#

Explanation:

The heat is transferred from the hot object to the cold water.

For the cold water, # Delta T_w=18ºC#

For the hot object #DeltaT_o=90ºC#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

The specific heat of water #C_w=4.186kJkg^-1K^-1#

Let, #C_o# be the specific heat of the object

The mass of the object is #m_o=0.020kg#

The mass of the water is #m_w=0.75kg#

#0.020*C_o*90=0.75*4.186*18#

#C_0=(0.75*4.186*18)/(0.020*90)#

#=31.4 kJkg^-1K^-1#