A 125 g piece of metal is heated to 288 0C and dropped into 85.0 g of water at 26.0 0C. If the final temperature of the water and metal is 58.0 0C, what is the specific heat of the metal in J/g 0C?

Answer 1

For this problem, we will need to use a couple of given equations and known values.

Q = m * c * ∆T

Q= Energy (Joules)

m= mass (grams)

c= specific heat capacity (Joules / g*C)

∆T= change in temperature, T_final - T_initial, Celsius

and algebraically moved to solve for "c";

c = Q / m*∆T

H₂0 has a specific heat capacity of 4.186 J/g°C;

c_H20= 4.186 J/g°C

We are told that both of our substances end up at the same temperature, so they are at equilibrium and this is a calorimetry problem.

Therefore we know that

Q₁ + Q₂ + Q₃..... = 0,

all of our substances' Q values will add to equal 0.

We have two substances, an unknown metal and water.

Now, let's find the Q value for each substance.

Q_H20 = mass * c * ∆T

1. H₂0 has a specific heat capacity (c) = 4.186 J/g°C.
2. mass (m) H₂0 = 85.0 g
3. ∆T = T_final - T_initial = 58.0 °C - 26.0 °C = +32.0°C

Q_H20 = mass * c * ∆T

Q_H20 = (85.0 g) (4.186 J/g°C)(+32.0°C)

Q_H20 = 11380.48 -->

3 sig figs

Q_H20= 1.14 x 10⁴

Q_metal = m * c * ∆T

1. mass = 125 g
2. c = unknown
3. ∆T = T_final - T_initial = 58.0 - 288 = -230°C

Q_metal = m * c * ∆T

= (125 g) (c) (-230°C)

= -28750(c)

----> 3 sig figs

Q_metal = -2.88 x 10⁴(c)

now, we can plug our Q values into our Q + Q = 0 equation;

Q_{metal +} Q_H20 = 0

-2.88 x 10⁴(c) + 1.14 x 10⁴ = 0

-2.88 x 10⁴(c) = -1.14 x 10⁴

(c) = (1.14 x 10⁴) / (2.88 x 10⁴)

*c = 3.96 x 10^-1 J/g°C, the specific heating capacity for our unknown metal.*