Question

A hot 123.3 g lump of an unknown substance initially at 164.2 °C is placed in 35.0 mL of water initially at 25.0 °C and the system is allowed to reach thermal equilibrium. The final temperature of the system is 48.6 °C. What is the identity of the unknown substance? Assume no heat is lost to the surroundings.

Answer 1

The unknown substance is rhodium.

Step-by-step explanation:

Heat lost by hot substance will be equal to heat gained by the water

`-Q_1=Q_2`

Mass of substance=
`m_1=123.3 g`

Specific heat capacity of substance=
`c_1=?`

Initial temperature of the substance =
`T_1=164.2^oC`

Final temperature of substance=
`T_2=T=48.6^oC`

`Q_1=m_1c_1* (T-T_1)`

Mass of water = M

Volume of water ,V= 35 ml

Density of water = d = 1.00 g/mL

`M=d* V=1.00 g/ml* 35 mL=35 g`

Mass of water=
`M = m_2=35 g`

Specific heat capacity of water=
`c_2=4.184 J/g^oC`

Initial temperature of the water =
`T_3=25.0^oC`

Final temperature of water =
`T_3=T=48.6^oC`

`Q_2=m_2c_2* (T-T_3)`

`-Q_1=Q_2`

`-(m_1c_1* (T-T_1))=m_2c_2* (T-T_3)`

On substituting all values:

we get,
`c_1 =0.242 J/g^oC`

The value of heat capacity of the substance is equal to that of rhodium metal.Hence, the unknown substance is rhodium.