Question

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system is 37.1°C, what is the specific heat capacity of the substance? Use 4.184 J / g °C for the specific heat capacity of water.

Answer 1

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$