An object with a mass of 120 g is dropped into 800 mL of water at 0^@C. If the object cools by 30 ^@C and the water warms by 5 ^@C, what is the specific heat of the material that the object is made of?

3 Answers
Jan 20, 2018

1.11 Calorie gm^-1 C^-1

Explanation:

Given,volume of water 800 ml,we know density of water is 1 (gm)/(cm^3) ,so it's mass will be 800 gm

Suppose,the specific heat of the object be s,then using law of thermal equilibrium we can write,

120×30×s = 800×5×1

Solving we get, s = 1.11 CGS units

Jan 20, 2018

Cp_o=(4.64J)/(g*^oC)

Explanation:

The important concept here is that "Heat lost=Heat gained"

Since the mass of the water is not given, we can derive it through its volume at a given temperature as provided. Knowing these, mass of the water can be calculated using the density formula; i.e.,

Density(rho)=("mass"(m))/("Volume"(V))
m=rhoxxV

where:
V=800ml " at " 0^oC https://hypertextbook.com/facts/2007/AllenMa.shtml

m=(0.9998g)/cancel(ml)xx800cancel(ml)
m=799.84g

When the object was submerged in the water, temperature of the water has increased to 5^o that signifies heat (Q) has been absorbed and that can be computed as follows:

Q=mCpDeltaT

where:

m_(H_20)=799.84g
T_i=0^oC
T_f=5^oC
Cp_w=(4.18J)/(g*^oC)

Q=799.84cancel(g)xx(4.18J)/cancel((g*oC))xx5cancel(*^oC)
Q=16,716.66J

Remember that "Heat lost" = "Heat gained"=16,716.66J. So that, heat capacity of the object (Cp_o) of this process can be computed with reference to this value of Q. Rearrange formula to isolate the Cp_o; that is,

Q=mCpDeltaT
Cp=(Q)/(mDeltaT)

where:

m=120g
DeltaT=30^oC

Cp=(16,716.66J)/(120gxx30^oC
Cp=(4.64J)/(g*^oC)

Jan 20, 2018

The specific heat is =4.65 kJkg^-1K^-1

Explanation:

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

For the cold water, Delta T_w=5ºC

For the object DeltaT_o=30ºC

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=0.12kg

The mass of the water is m_w=0.800kg

0.12*C_o*30=0.800*4.186*5

C_o=(0.800*4.186*5)/(0.12*30)

=4.65 kJkg^-1K^-1