Solution.
If during the burning of 1 gram of acetylene 48.2 kJ is released, then during the burning of 0.75 grams it is released:
Q=10.75×48.2Q = 36.15 kJ
Q=q×Δtq = 1.117 kJ/°C
Δt=qQΔt=t2−t1Δt=1.11736.15
Δt=32.36
t1=t2−Δtt1 = 22.14 °C
Answer:
t1 = 22.14 °C
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