Question

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?

Answer 1

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

Answer 2

`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)`

Answer 3

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`