what is the initial temperature of the mercury?

Question

A sample of liquid mercury is heated with an electrical coil. If 32.3 Joules of energy are added to a 12.5 gram sample and the final temperature is 38.8°C, what is the initial temperature of the mercury?

Answer: °C.

Answer 1

Hi Mawi!

This is a problem that's pretty common in enthalpy-style calculations. It's one of the most commonly misunderstood concepts, because in common speech, we tend to use heat and temperature interchangably. This is the way that they're truly related.

In our problem, we're taking a sample of liquid mercury and adding heat to it, consequently changing its temperature. One thing that is needed for this problem is the specific heat capacity of liquid mercury, which I looked up -- it's 0.14 J/g*K.

We're going to use the formula ΔH = m*c*ΔT

In which ΔH stands for enthalpy (the amount of energy we put into the system), m is the mass (in grams), c is the specific heat capacity of liquid mercury (0.14 J/g*K), and ΔT, which is the change in temperature (in K...or °C, since it's a difference).

ΔH = m*c*ΔT

32.3 J = 12.5g * 0.14 J/g*K * ΔT

So if we solve for ΔT, we get...ΔT = 18.46K (which is the same as 18.46°C)

Since we added energy to the system, we know that the temperature goes up -- so if our final temperature was 38.8°C, our initial temperature would be...

Final temperature - change in temperature = initial temperature

38.8°C - 18.46°C = 20.34°C

Hope that helps!

-Niral

Answer 2

To solve the problem, you need to use the formula...

q = mcΔt

q = heat absorbed (in joules)

m = mass (in grams)

c = specific heat capacity (of mercury)

Δt = change in temperature (in ^oC)

In this case, q = 32.3, m = 12.5, and c = 0.14 (according to google). Therefore, you just need to solve for Δt.

32.3 = (12.5)(0.14)(Δt)

32.3 / [(12.5)(0.14)] = Δt

18.46 = Δt

However, you aren't finished here because you still need to find the initial temperature, and all we have just found was the change in temperature. Therefore you subtract your change in temperature from your final temperature to get your initial temperature as follows...

38.8 - 18.46 = 20.34^oC