When 72 grams of a metal at 97.0oC is added to 100.0 grams of water at 25oC, the final temperature is 29.1oC. What is the specific heat capacity (in J/g.oC) of the metal?
AW C = 12 g/mol ; AW O = 16 g/mol
HATDOG
Well, it seems like the metal is quite the chill master! It managed to cool down that water from 97.0oC to 29.1oC. Impressive!
To calculate the specific heat capacity of the metal, we can use the formula:
q = mcΔT
Where q is the heat exchanged, m is the mass of the substance (metal or water in this case), c is the specific heat capacity, and ΔT is the change in temperature.
For the water, the heat exchanged is given by:
q_water = mc_waterΔT_water
Since there is no phase change happening in the water, we can use its specific heat capacity of 4.18 J/g.oC.
Now, for the metal, the heat exchanged is given by:
q_metal = mc_metalΔT_metal
Since we're trying to find the specific heat capacity of the metal, let's represent it as c_metal.
q_metal = mc_metalΔT_metal = -q_water = -mc_waterΔT_water
Dividing both ends of the equation by the mass and change in temperature, we get:
c_metal = -c_water(ΔT_water/ΔT_metal)
Plugging in the values we have:
c_metal = -(4.18 J/g.oC)(29.1oC - 25oC)/(97.0oC - 29.1oC)
Do the math and you'll find the specific heat capacity of this metal, which is likely making all the other metals jealous of its coolness!
To find the specific heat capacity of the metal, we can use the formula:
q = mcΔT
Where:
- q is the heat flow (in Joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity (in J/g.oC)
- ΔT is the change in temperature (in oC)
First, we need to calculate the heat flow between the metal and water using the formula:
q1 = m1c1ΔT1
Where:
- q1 is the heat flow between the metal and water
- m1 is the mass of the metal (in grams)
- c1 is the specific heat capacity of the metal (which we want to find)
- ΔT1 is the change in temperature between the initial and final temperatures of the system
We can rearrange the equation to solve for c1:
c1 = q1 / (m1 * ΔT1)
Next, we calculate the heat flow between the water and metal:
q2 = m2c2ΔT2
Where:
- q2 is the heat flow between the water and metal
- m2 is the mass of the water (in grams)
- c2 is the specific heat capacity of water (which is approximately 4.18 J/g.oC)
- ΔT2 is the change in temperature between the initial and final temperatures of the system
Since the heat lost by the metal is equal to the heat gained by the water, we have:
q1 = -q2 (since heat lost = -1 * heat gained)
Now we can substitute the values into the equations:
q1 = m1c1ΔT1
q2 = m2c2ΔT2
q1 = -q2
We have the following information:
m1 = 72 g (mass of the metal)
m2 = 100 g (mass of the water)
c2 = 4.18 J/g.oC (specific heat capacity of water)
ΔT1 = (29.1 - 97) oC (change in temperature for the metal)
ΔT2 = (29.1 - 25) oC (change in temperature for the water)
Now we can calculate q1 and q2:
q1 = m1c1ΔT1
q2 = m2c2ΔT2
q1 = -q2
Finally, we can solve for c1:
c1 = q1 / (m1 * ΔT1)
By substituting the known values into the equations and performing the necessary calculations, we can find the specific heat capacity of the metal.
Sjs
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute the numbers and solve for specific heat metal.