Question

A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5°C. Use these data to determine the specific heat of the metal.

Answer 1

Answer:

$s = 382.45 J/kg C$

Explanation:

Here by energy equivalence we can say that energy given by the metal piece is same as the energy absorbed by the water

so here we have

$m_1s_2\Delta T_1 = m_2s_2 \Delta T_2$

here we know that

$m_1 = 60 mL = 0.060 kg$

$s_1 = 4186 J/kg C$

$\Delta T_1 = 28.5 - 22$

$m_2 = 59.7 g = 0.0597 kg$

$\Delta T_2 = 100 - 28.5$

$0.060(4186)(28.5 - 22) = (0.0597)(s)(100 - 28.5)$

$1632.54 = 4.27 s$

$s = 382.45 J/kg C$