The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
<h3>What is the specific heat capacity?</h3>
The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
Given data;
Mass of liquid sample of Alcohol m₁ = 200-gram
The temperature of alcohol, T₁ = -6°C.
Mass of liquid sample of water m₂ = 400-gram
The temperature of the water, T₂= 20°C.
The specific heat capacity of the alcohol, S₁=?
The specific heat capacity of water is, S₂=4.19 kJ/kg.°C
As we know that;
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![\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C](https://tex.z-dn.net/?f=%5Crm%20Q_%7Bgain%7D%3D%20Q%7Bloss%7D%20%5C%5C%5C%5C%20Q_%7Balcohol%7D%20%3DQ_%7Bwater%7D%20%5C%5C%5C%5C%5C%20m_1s_1%5Ctriangle%20T_1%20%3D%20m_2S_2%20%5Ctriangle%20S_2%20%5C%5C%5C%5C%20200%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20S_1%20%5B%20%2812-%28-6%29%20%5D%20%3D%2040%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%204.19%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20%2820-12%29%5C%5C%5C%5CS_1%20%3D%202%20%5Ctimes%204.19%20%5Ctimes%2010%5E3%20%5Ctimes%20%5Cfrac%20%7B8%7D%7B18%7D%20%5C%5C%5C%5C%20S_1%20%3D%203.72%20%20%5C%20kJ%20%2Fkg%20%5E0%20C)
Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
To learn more about the specific heat capacity, refer to the link brainly.com/question/2530523.
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