another calorimetry experiment is performed in which a 500. gram peice of gray metal at 67 degrees celsius is added to 493 grams of water. the water's temperature, originally at 17 degrees celsius, changes to 22 degrees celsius. What is the specific heat of the gray metal? I know that you use the formula q=m*c* delta T. but how do i do this problem
q for metal + q for water = 0
q for metal is m*c*delta t.
q for water is m*c*delta t.
The only unknown is c for the metal.
is the answer .9
No it isn't. You may not be substituting correctly. For delta T, make sure it is (Tfinal-Tinitial). That will make the signs right. I found an answer less than 1 J/g*C.
THANKSS DOC.
You're welcome.
To calculate the specific heat of the gray metal, you can use the formula q = m * c * ΔT, where:
- q represents the amount of heat transferred
- m represents the mass of the substance (either the metal or the water in this case)
- c represents the specific heat capacity of the substance
- ΔT represents the change in temperature
In this scenario, the heat transferred from the metal to the water can be calculated using the formula:
q metal + q water = 0
The heat transferred from the metal to the water, q metal, can be expressed as:
q metal = -q water
Let's calculate the heat transferred from the water, q water, first:
q water = m water * c water * ΔT water
Given:
m water = 493 grams (mass of water)
c water = 4.18 J/g°C (specific heat capacity of water)
ΔT water = (final temperature of water - initial temperature of water)
= (22°C - 17°C)
Now, let's find the heat transferred from the metal, q metal:
q metal = -q water
= -m water * c water * ΔT water
Since we have the heat transferred per gram, we need to multiply the mass of the metal by its specific heat capacity (c metal) to find the heat transferred from the metal:
q metal = -m metal * c metal * ΔT water
Given:
m metal = 500 grams (mass of the gray metal)
ΔT water = (22°C - 17°C)
Now, rearranging the equation to solve for the specific heat capacity of the metal (c metal):
c metal = -q metal / (m metal * ΔT water)
Plugging in the given values:
c metal = -q metal / (500 g * (22°C - 17°C))
Calculating the change in heat, q metal:
q metal = -q water
= -m water * c water * ΔT water
= -493 g * 4.18 J/g°C * (22°C - 17°C)
Substituting the value of q metal into the equation for c metal:
c metal = -(-493 g * 4.18 J/g°C * (22°C - 17°C)) / (500 g * (22°C - 17°C))
Simplifying the equation:
c metal = 5.122 J/g°C
Therefore, the specific heat of the gray metal is approximately 5.122 J/g°C.