HELP!!! An unknown metal having a mass of 14.00g is dropped into 11.00mL of water at 20.0 degrees Celcius. The starting temperature of the chunk of metal is 100.0C. The final temperature of the water and metal is 32.5C. What is the specific heat of the metal?
The trick is that the heat lost from the metal is equal to the heat gained by the water.
So q lost = f gained. Since the metal went from 100.0C to 32.5C I'd say it lost. The water went from 20.0C to 32.5C, so it gained. "q" then represents the metal and "f" represents the water.
Now apply the following formula to solve for "q" and for "f":
(mass)(temp change)(specific heat).
What it should look like is this, (mass)(temp change)(specific heat) = (mass)(temp change)(specific heat).
Now plug in the values.
Your metal has a mass of 14.00g. The water's mass is 11,000g (which is 11mL).
Temp change of the metal was 100.0C down to 32.5C, so 67.5C. Temp change for the water was 20.0C up to 32.5C, so 12.5C.
Specific heat of your metal is "x" which we're solving for. Specific heat of water is 1.00cal/g x C.
I can't write out the equation on this web page. Fill in all the parenthesis making one set equal to the other. And then solve for x.
To find the specific heat of the metal, we can use the equation:
q = mcΔT
where:
q is the heat gained or lost
m is the mass of the metal
c is the specific heat capacity of the metal
ΔT is the change in temperature
We need to calculate the heat gained or lost by the water using the equation:
q = mcΔT
where:
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's find the heat gained or lost by the water using the given information.
The mass of the water is given as 11.00 mL. However, to use this mass in the equation, we need to convert it to grams. Since 1 mL of water has a mass of 1 gram, the mass of the water is also 11.00 grams.
The specific heat capacity of water, c, is approximately 4.18 J/g·°C.
Now we can calculate the heat gained or lost by the water using the equation:
q = mcΔT
q = (11.00 g)(4.18 J/g·°C)(32.5 °C - 20.0 °C)
After performing the calculation, we obtain:
q = 550.82 J
Next, let's calculate the heat gained or lost by the metal using the equation for q:
q = mcΔT
Since q is the same for both the water and the metal, we can set up an equation:
q (metal) = q (water)
Let's rearrange the equation to solve for c:
c = q (metal) / (m (metal) ΔT (metal))
We are given the mass of the metal as 14.00 g and the change in temperature as (100.0 °C - 32.5 °C).
Plugging the values into the equation, we have:
c = 550.82 J / (14.00 g)(100.0 °C - 32.5 °C)
After performing the calculation, we obtain:
c ≈ 0.717 J/g·°C
Therefore, the specific heat of the metal is approximately 0.717 J/g·°C.