I post this question two time before but the degree sign is wrong so I post this one with everything corrected for the final time. Thank upon who answer this question on the other 2 same question before. Calculate the final temperature that results from mixing 245 grams of cobalt, specific heat 0.446 J/g degree C, at 142 degree C with 106 grams of water at 24.8 degree C.
heat lost by Co + heat gained by H2O = 0
[mass Co x specific heat Co x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
Subsitute and solve for Tf which is the only unknown in the equation.
To calculate the final temperature after mixing cobalt and water, we can use the principle of conservation of energy. The heat gained by the cobalt will be equal to the heat lost by the water. The formula to calculate the heat gained or lost is:
q = m * c * ΔT
Where:
q is the heat gained or lost
m is the mass
c is the specific heat
ΔT is the change in temperature
For the cobalt, the heat gained (q1) can be calculated as:
q1 = m1 * c1 * ΔT1
= 245 g * 0.446 J/g°C * (Tf - 142°C)
For the water, the heat lost (q2) can be calculated as:
q2 = m2 * c2 * ΔT2
= 106 g * 4.18 J/g°C * (Tf - 24.8°C)
Since the heat gained by the cobalt is equal to the heat lost by the water, we can set q1 equal to q2:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Plugging in the values:
245 g * 0.446 J/g°C * (Tf - 142°C) = 106 g * 4.18 J/g°C * (Tf - 24.8°C)
Now, we can solve for Tf.
245 * 0.446 * (Tf - 142) = 106 * 4.18 * (Tf - 24.8)
Multiply:
109.37 * (Tf - 142) = 444.68 * (Tf - 24.8)
Distribute:
109.37 * Tf - 109.37 * 142 = 444.68 * Tf - 444.68 * 24.8
Simplify:
109.37Tf - 15470.54 = 444.68Tf - 11024.864
Move terms around:
109.37Tf - 444.68Tf = -11024.864 + 15470.54
Combine like terms:
-335.31Tf = 4436.676
Divide both sides by -335.31:
Tf = 4436.676 / -335.31
Tf ≈ -13.2°C
Therefore, the final temperature after mixing the cobalt and water will be approximately -13.2°C.