Answer:
119.6 J/Kg°C
Explanation:
Data obtained from the question include:
Mass of substance (ms) = 170 g
Initial temperature of substance (Ts) = 120 °C
Volume of water = 200 mL
Initial temperature of water (Ts) = 10 °C
Temperature of the mixture (T2) = 12.6 °C
Density of water = 1 g/mL
Specific heat capacity of water (Cw) = 4200J/Kg°C
Specific heat capacity of substance (Cs) =..?
Next, we shall determine the mass of water. This can be obtained as follow:
Volume of water = 200 mL
Density of water = 1 g/mL
Mass of water =..?
Density = mass /volume
1 = mass /200
Cross multiply
Mass of water = 1 x 200
Mass of water = 200 g
Convert 200 g of water to Kg
Mass of water = 200/1000 0.2 Kg
Mass of water = 0.2 Kg
Now, we obtained the specific heat capacity of the substance using the following formula:
MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0
Mass of water = 0.2 Kg
Initial temperature of water (Ts) = 10 °C
Specific heat capacity of water (Cw) = 4200J/Kg°C
Temperature of the mixture (T2) = 12.6 °C
Mass of substance (ms) = 170 g = 170/1000 = 0.17 Kg
Initial temperature of substance (Ts) = 120 °C
Specific heat capacity of substance (Cs) =..?
MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0
0.2× 4200(12.6 – 10) + 0.17×Cs×(12.6 – 120) = 0
840(2.6) + 0.17Cs(– 107.4) = 0
2184 – 18.258Cs = 0
Rearrange
2184 = 18.258Cs
Divide both side by the coefficient of Cs i.e 18258
Cs = 2184/18.258
Cs = 119.6 J/Kg°C
Therefore, the specific heat capacity of the substance is 119.6 J/Kg°C
radical and
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