Answer:
The answer to your question is C = 0.037 cal/g°C
Explanation:
Data
mass of the wire = m = 237 g
temperature 1 = T1 = 25°C
temperature 2 = T2 = 107°C
Heat = Q = 722 cal
Specific heat = C
Process
1.- Write the formula to find the specific heat
Q = mC(T2 - T1)
-Solve for C
C = Q / m(T2 - T1)
2.- Substitution
C = 722 / 237(107 - 25)
3.- Simplification
C = 722 / 237(82)
C = 722 / 19434
4.- Result
C = 0.037 cal/g°C
I think it might be a closed system
Boat p
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C2H6 OR C6H12O6 THE ANSWER
