Answer:
0.4589J/g°C
Explanation:
Heat energy = 1050J
Mass of iron = 220.0g
Initial temperature (T1) = 10°C
Final temperature (T2) = 20.4°C
Heat energy = mc∇T
Q = mc∇T
Q = heat energy
M = mass of the substance
C = specific heat capacity of the substance
∇T = change in temperature = T2 - T1
Q = m×c×(T2 - T1)
1050 = 220 × c ×(20.4 - 10)
1050 = 220c × (10.4)
1050 = 2288c
c = 1050 / 2288
C = 0.4589J/g°C
The specific heat capacity of iron is 0.4589J/g°C
Answer:
0.4589 J g^-1 K^-1 IS THE SPECIFIC HEAT OF IRON.
Explanation:
Specific heat of iron can be calculated using the heat of reaction formulae
Heat = mass * specific heat * ( T2-T1)
H = m c (T2-T1)
c = H / m * (T2-T1)
H = 1050 J
m = 220.0 g
T2 = 20.4 ° C = 20.4 + 273 = 293.4 K
T1 = 10.0 °C = 10 + 273 K = 283 K
so therefore:
c = 1050 / 220 * ( 293.4 - 283) K
c = 1050 / 220 * 10.4
c = 1050 / 2288
c = 0.4589 J g^-1 K^-1
The specific heat capacity of iron is 0.4589 J g^-1 K^-1
