When 80.0 grams of a certain metal at 90.0 °C was mixed with 100.0 grams of water at 30.0 °C, the final equilibrium temperature of the mixture was 36.0 °C. What is the specific heat (J/g°C) of the metal?
I suppose I would use this formula:
q= mcÄt
m= mass(g)
q= enthalpy change (j)
c= specifice heat capacity(J/g°C)
Ät= temperature change(°C)
not quite sure how to approach this question!?!?!?!
The sum of the heats added is zero (one is lost, so it is negative).
Heataddedwater + heataddedmetal=0
mw*cw*(Tf-Tiw) + mm*Cm*(Tf-Tim)=0
solve for Cm, the specific heat of the metal.
Well, isn't this a hot topic! Let's dive in and solve this equation, shall we?
We can start by plugging in the given values into the equation:
mw * cw * (Tf - Tiw) + mm * Cm * (Tf - Tim) = 0
Here's the breakdown of what each term represents:
mw - mass of water (100.0 grams)
cw - specific heat capacity of water (4.18 J/g°C)
Tf - final equilibrium temperature of the mixture (36.0 °C)
Tiw - initial temperature of water (30.0 °C)
mm - mass of metal (80.0 grams)
Cm - specific heat capacity of the metal (what we're solving for!)
Tim - initial temperature of the metal (90.0 °C)
Now let's substitute these values into the equation and simplify:
100.0 * 4.18 * (36.0 - 30.0) + 80.0 * Cm * (36.0 - 90.0) = 0
Now solve for Cm by solving this equation. Let's crunch some numbers!
(100.0 * 4.18 * 6) + (80.0 * Cm * -54.0) = 0
2498.8 - 4320.0 * Cm = 0
-1821.2 * Cm = -2498.8
Cm = -2498.8 / -1821.2
Cm ≈ 1.37 J/g°C
So, it looks like the specific heat capacity of the metal is approximately 1.37 J/g°C. Phew, that was heated, but we made it through!
To find the specific heat (Cm) of the metal, we can use the equation:
mw * cw * (Tf - Tiw) + mm * Cm * (Tf - Tim) = 0
Where:
mw = mass of water (100.0 g)
cw = specific heat capacity of water (4.184 J/g°C)
Tf = final equilibrium temperature (36.0 °C)
Tiw = initial temperature of water (30.0 °C)
mm = mass of metal (80.0 g)
Cm = specific heat capacity of the metal (unknown)
Tim = initial temperature of the metal (90.0 °C)
Substituting the given values into the equation, we get:
(100.0 g) * (4.184 J/g°C) * (36.0 °C - 30.0 °C) + (80.0 g) * Cm * (36.0 °C - 90.0 °C) = 0
Simplifying this equation, we have:
(100.0 g) * (4.184 J/g°C) * (6.0 °C) + (80.0 g) * Cm * (-54.0 °C) = 0
(2510.4 J) - (4320.0 g * Cm) = 0
-4320.0 g * Cm = -2510.4 J
Dividing both sides of the equation by -4320.0 g, we get:
Cm = -2510.4 J / -4320.0 g
Cm ≈ 0.581 J/g°C
Therefore, the specific heat of the metal is approximately 0.581 J/g°C.
To solve this problem, you can use the equation:
q = mcΔT
Where:
q refers to the heat transferred (in joules),
m refers to the mass of the substance (in grams),
c refers to the specific heat capacity of the substance (in J/g°C), and
ΔT refers to the change in temperature (in °C).
In this case, you have two separate substances: the metal and the water. The heat added to the water will be equal in magnitude, but opposite in sign, to the heat lost by the metal. So, you can set up the equation as follows:
mw * cw * (Tf - Tiw) + mm * cm * (Tf - Tim) = 0
where:
mw refers to the mass of water (100.0 grams)
cw refers to the specific heat capacity of water (4.18 J/g°C)
Tiw refers to the initial temperature of water (30.0 °C)
mm refers to the mass of the metal (80.0 grams)
cm refers to the specific heat capacity of the metal (what we are trying to find)
Tim refers to the initial temperature of the metal (90.0 °C)
Tf refers to the final equilibrium temperature (36.0 °C)
Now you can plug in the values into the equation and solve for cm:
(100.0 g) * (4.18 J/g°C) * (36.0 °C - 30.0 °C) + (80.0 g) * cm * (36.0 °C - 90.0 °C) = 0
Simplifying this equation will allow you to isolate cm and find the specific heat capacity of the metal.