Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
Answer:
The answer to your question is Q = 18702.5 J
Explanation:
Data
mass of water = m = 447 g
Cp = 4.184 J/g°C
Temperature 1 = T1 = 25°C
Temperature 2 = T2 = 35°C
Heat = Q = ? Joules
Process
1.- Write the formula to calculate heat
Q = mCp(T2 - T1)
2.- Substitution
Q = (447)(4.184)(35 - 25)
3.- Simplification
Q = (447)(4.184)(10)
4.- Result
Q = 18702.5 J
Answer: C.)
Explanation:
i got it right on a unit test!
but it might be something else if there arranged different!
sorry!
I believe the answer is B. PO4-3
