Can someone walk me through the steps on how to solve using these equations? Thank yoouuuu~~
1.11g of methane CH4 is burned in a bomb calorimeter containing 1000 grams of water. The initial temperature of water is 24.85oC. The specific heat of water is 4.184 J/g oC. The heat capacity of the calorimeter is 695 J/ oC. After the reaction the final temperature of the water is 35.65oC.
Using the formula water q =m•c• ΔT,calculate the heat absorbed by the water
Using the formula cal cal q =C • ΔT,calculate the heat absorbed by the
calorimeter.
I don't get it. You're given the formula(s) to use. All you need to do is to substitute the values in the problem.
For the water.
q = mc(Tfinal-Tinitial)
q = 1000g x 4.184 J/g*C x (35.65-24.85)= ?
q is + so heat is absorbed.
For the cal.
q = C*(Tfinal-Tinitial)
q = 695 J/C x (35.65-24.85) = ?
q is + so heat is absorbed.
To solve this problem, we need to follow these steps:
Step 1: Determine the heat absorbed by the water.
The formula to calculate the heat absorbed by the water is q = m • c • ΔT, where q is the heat absorbed, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature (final temperature - initial temperature).
Given values:
m (mass of water) = 1000 grams
c (specific heat of water) = 4.184 J/g°C
ΔT (change in temperature) = 35.65°C - 24.85°C = 10.8°C
Calculating the heat absorbed by the water:
q = 1000 grams • 4.184 J/g°C • 10.8°C
q = 45,283.2 J
Therefore, the heat absorbed by the water is 45,283.2 J.
Step 2: Determine the heat absorbed by the calorimeter.
The formula to calculate the heat absorbed by the calorimeter is qc = C • ΔT, where qc is the heat absorbed, C is the heat capacity of the calorimeter, and ΔT is the change in temperature (final temperature - initial temperature).
Given value:
C (heat capacity of the calorimeter) = 695 J/°C
ΔT (change in temperature) = 35.65°C - 24.85°C = 10.8°C
Calculating the heat absorbed by the calorimeter:
qc = 695 J/°C • 10.8°C
qc = 7,506 J
Therefore, the heat absorbed by the calorimeter is 7,506 J.
So, the heat absorbed by the water is 45,283.2 J, and the heat absorbed by the calorimeter is 7,506 J.