Answer:
Specific heat capacity of iron (C) = 0.3 J/GK
Explanation:
Given:
Mass of sample (m) = 125 gram
Change in heat (ΔT) = 20°C - 0°C = 20°C
Absorbed energy (ΔE) = 750 J
Find:
Specific heat capacity of iron (C) = ?
Computation:
⇒ Specific heat capacity of iron (C) = ΔE / [mΔT]
⇒ Specific heat capacity of iron (C) = 750 / [125 × 20]
⇒ Specific heat capacity of iron (C) = 750 / [2,500]
⇒ Specific heat capacity of iron (C) = 0.3 J/GK
1 mol Cu - 2 mol Ag
x - 0.854 mol Ag
x=0.854*1/2=0.427
0.427 mol Cu
C + 273.15 = temp. in Kelvins.
25° C would be 298.15 k.
