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

Answer:

The unknown substance is rhodium.

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= $T_2=T=48.6^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water = M

Volume of water ,V= 35 ml

Density of water = d = 1.00 g/mL

$M=d\times V=1.00 g/ml\times 35 mL=35 g$

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 = $T_3=T=48.6^oC$

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

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.