An object with a mass of #10 g# is dropped into #240 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

2 Answers
Feb 28, 2018

Here,the heat energy liberated by the object will be taken up by the water as a result of which,both will come to a thermal equilibrium.

So,heat energy liberated by the object is #10*s* 120 # Calorie (using #H= msd theta#,where, #m# is the mass of the object, #s# is its specific heat and #d theta# is the change in temperature)

Now,heat energy absorbed by water is #240*1*1*8# calorie (where, mass of water = volume* density, and #1 ml =1 cm^3# and in CGS, density of water is #1 g/(cm^3)# and specific heat of water is #1 C g^-1 @C ^-1#)

So,equating both we get,

# 1200s =1920#

so, #s= 1.6 C g^-1 @C^-1#

Feb 28, 2018

Answer:

The specific heat is #=6.76.7 kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=8ºC#

For the object #DeltaT_o=120ºC#

# 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=0.010kg#

The mass of the water is #m_w=0.24kg#

#0.010*C_o*120=0.24*4.186*8#

#C_o=(0.24*4.186*8)/(0.010*120)#

#=6.7 kJkg^-1K^-1#