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A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 12°C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.19 kJ/kg · °C. 3.14 kJ/kg \xe2\x88\x99 °C 4.14 kJ/kg \xe2\x88\x99 °C 3.72 kJ/kg \xe2\x88\x99 °C 4.88 kJ/kg \xe2\x88\x99 °C

Sagot :

The specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

What is the specific heat capacity?

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;

[tex]\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[/tex]

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 https://brainly.com/question/2530523.

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