Question

A 10.0 kg piece of metal at 50.0oC is placed in 1.00 kg of water at 10.0oC. The metal and water come to the same temperature of 31.4oC. What is the specific heat of the metal?

Answer 1

c_m = 0.4814 J/g°C

Step-by-step explanation:

From thermo equilibrium conditions, we now that;

Heat lost by the Metal = Heat gained by the Water

Thus;

-Q_m = Q_w

Q_m = m_m × c_m × Δt_m

Also,

Q_w = m_w × c_w × Δt_w

We are given;

Mass of metal; m_m = 10 kg

Initial temperature of the metal; t_mi = 50°C

Change in temperature of metal; Δt_m = (31.4 - 50) = -18.6 °C

Mass of water; m_w = 1 kg

Initial temperature of the water; t_wi = 10°C

Change in temperature of water; Δt_w = 31.4°C - 10°C = 21.4 °C

Specific heat capacity of water; c_w = 4.184 J/g°C

We are looking for c_m which is the specific heat capacity of the metal.

Now, from -Q_m = Q_w;

-(m_m × c_m × Δt_m) = (m_w × c_w × Δt_w)

Let's make c_m the subject;

c_m = (m_w × c_w × Δt_w)/(-m_m × Δt_m)

Plugging in the relevant values;

c_m = (1 × 4.184 × 21.4)/(-10 × -18.6)

c_m = 0.4814 J/g°C