Answer:
0.035 J/g°C
Explanation:
From the question given above, the following data were obtained:
Heat (Q) absorbed = 2500 J
Mass (M) = 1.2 Kg
Initial Temperature (T₁) = 10 °C
Final Temperature (T₂) = 70 °C
Specific heat capacity (C) =?
Next, we shall determine the change in temperature. This can be obtained as follow:
Initial Temperature (T₁) = 10 °C
Final Temperature (T₂) = 70 °C
Change in temperature (ΔT) =
ΔT = T₂ – T₁
ΔT = 70 – 10
ΔT = 60 °C
Thus, the change in the temperature of the substance is 60 °C
Next, we shall convert 1.2 Kg to grams (g). This can be obtained as follow:
1 Kg = 1000 g
Therefore,
1.2 Kg = 1.2 Kg × 1000 g / 1 Kg
1.2 Kg = 1200 g
Thus, 1.2 Kg is equivalent to 1200 g.
Finally, we shall determine the specific heat capacity of substance. This can be obtained as follow:
Heat (Q) absorbed = 2500 J
Mass (M) = 1200 g
Change in temperature (ΔT) = 60 °C
Specific heat capacity (C) =?
Q = MCΔT
2500 = 1200 × C × 60
2500 = 72000 × C
Divide both side by 72000
C = 2500 / 72000
C = 0.035 J/g°C
Therefore, the specific heat capacity of the substance is 0.035 J/g°C
