Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
You are correct, but you needn't worry about the signs so much. Just remember that the negative sign is used to denote a loss of energy; since the water is hotter, it will be losing energy (-Q) and the iron will gain energy (Q). Now, we substitute the values:
-149.3 * 4.184 * (T - 95) = 412 * 0.44 * (T - 5)
Solving this equation for T,
T = 74.8 °C
It is c I hope I helped out with this question!.
A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)