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?

c_m = 0.4814 J/g°C

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

mention the spill to the teacher clean the area and throw the broken pieces away

explanation:

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.