Question

A silver cube with an edge length of 2.28 cm and a gold cube with an edge length of 2.75 cm are both heated to 82.8 ∘C and placed in 111.5 mL of water at 20.2 ∘C . What is the final temperature of the water when thermal equilibrium is reached?

Answer 1

Solution :

Given :

length of silver cube = 2.28 cm

length of gold cube = 2.75 cm

Initial temperature = 82.8°C

Volume of silver cube is

Volume $=(\text{edge length})^3$

               =    $(2.28)^3$

                $= 11.8 \ cm^3$

mass of silver cube

Mass, $m_s = \text{density} \times \text{volume}$

                = 10.5 x 11.8

                = 123.9 g

Similarly, the volume of gold cube is

Volume $=(\text{edge length})^3$

               =    $(2.75)^3$

                $= 20.79 \ cm^3$

mass of gold cube

Mass, $m_g = \text{density} \times \text{volume}$

                = 19.3 x 20.79

                = 401.247 g

Now

heat lost by silver and gold cube = heat gained by water

∴$m_g .c_g \Delta T + m_s .c_s \Delta T = m_w.c_w \Delta T$

$m_g .c_g (82.8- T) + m_s .c_s (82.8- T) = m_w.c_w (T - 20.2)$

$401.247 \times 0.1264 (82.8- T) + 123.9 \times 0.2386 (82.8- T) = 111.5 \times 4.184 (T - 20.2)$

Now solving the equation

$50.71 (82.8- T) + 29.56 (82.8- T) = 466.51 (T - 20.2)$

Final temperature, T = 31.27°C