Answer:
0.417 J/gºC
Explanation:
From the question given above, the following data were obtained:
Heat (Q) absorbed = 2500 J
Mass (M) of substance = 100 g
Initial temperature (T1) = 10 °C
Final temperature (T2) = 70 °C
Specific heat capacity (C) =?
Next, we shall determine the change in temperature (ΔT). This can be obtained as follow:
Initial temperature (T1) = 10 °C
Final temperature (T2) = 70 °C
Change in temperature (ΔT) =?
Change in temperature (ΔT) = T2 – T1
Change in temperature (ΔT) = 70 – 10
Change in temperature (ΔT) = 60 °C
Finally, we shall determine the specific heat capacity of the substance as follow:
Heat (Q) absorbed = 2500 J
Mass (M) of substance = 100 g
Change in temperature (ΔT) = 60 °C
Specific heat capacity (C) =?
Q = MCΔT
2500 = 100 × C × 60
2500 = 6000 × C
Divide both side by 6000
C = 2500 / 6000
C = 0.417 J/gºC
Therefore, the specific heat capacity of substance is 0.417 J/gºC
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