A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system is 37.1°C, what is the specific heat capacity of the substance? Use 4.184 J / g °C for the specific heat capacity of water. A. 368 J / g °C B. 13.1 J / g °C C. 0.368 J / g °C D. 0.0112 J / g °C
Question
A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system is 37.1°C, what is the specific heat capacity of the substance? Use 4.184 J / g °C for the specific heat capacity of water. A. 368 J / g °C B. 13.1 J / g °C C. 0.368 J / g °C D. 0.0112 J / g °C
Solution
To solve this problem, we need to use the formula for heat transfer:
q = mcΔT
where: q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
We know that the heat gained by the water is equal to the heat lost by the substance. Therefore, we can set up the following equation:
m_water * c_water * ΔT_water = m_substance * c_substance * ΔT_substance
Substituting the given values:
1,150 g * 4.184 J/g°C * (37.1°C - 25.0°C) = 75.0 g * c_substance * (96.5°C - 37.1°C)
Solving for c_substance, we get:
c_substance = (1,150 g * 4.184 J/g°C * 12.1°C) / (75.0 g * 59.4°C)
After calculating the above expression, we find that the specific heat capacity of the substance is approximately 0.368 J/g°C. Therefore, the correct answer is C. 0.368 J/g°C.
Similar Questions
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