Answer to Question #194724 in Physics for John

Question

 A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.

Answer 1

Let's first find the quantity of heat lost by the water:

$$Q_w=m_wc_w(T_f-T_i),$$ $$Q_w=50.0\ g\cdot4.18\ \dfrac{J}{g\cdot \!^{\circ}C}\cdot(87.1^{\circ}C-88.6^{\circ}C)=-313.5\ J.$$

The sign minus means that the heat is lost by the water.

The metal is gained the quantity of heat lost by the water:

$$Q_m=Q_w=+313.5\ J.$$

Here, the sign plus means that the heat is gained by the metal.

Then, we can write the formula for the quantity of heat gained by the metal:

$$Q_m=m_mc_m(T_f-T_i).$$

From this formula we can find the specific heat capacity of the unknown metal:

$$c_m=\dfrac{Q_m}{m_m(T_f-T_i)},$$ $$c_m=\dfrac{313.5\ J}{12.9\ g\cdot(87.1^{\circ}C-26.5^{\circ}C)}=0.4\ \dfrac{J}{g\cdot \!^{\circ}C}.$$