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A 54.8 g sample of quartz, which has a specific heat capacity of 0.730 J-g¹C¹, is put into a calorimeter (see sketch at right)

that contains 200.0 g of water. The temperature of the water starts off at 23.0 °C. When the temperature of the water stops
changing it's 26.4 °C. The pressure remains constant at 1 atm.
Calculate the initial temperature of the quartz sample. Be sure your answer is rounded to 2 significant digits.

1 Answer

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Answer:

The initial temperature of the quartz sample was 16.3 °C.

Step-by-step explanation:

The heat lost by the quartz sample is equal to the heat gained by the water and the calorimeter. Therefore:

q_quartz = q_water + q_calorimeter

The heat lost by the quartz sample can be calculated using the equation:

q_quartz = m_quartz × c_quartz × ΔT_quartz

where m_quartz is the mass of the quartz sample, c_quartz is the specific heat capacity of quartz, and ΔT_quartz is the change in temperature of the quartz sample.

The heat gained by the water and calorimeter can be calculated using the equation:

q_water + q_calorimeter = (m_water + m_calorimeter) × c_water × ΔT_water

where m_water is the mass of the water, m_calorimeter is the mass of the calorimeter, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water and calorimeter.

Since the pressure is constant and no work is being done, we can assume that the total heat gained by the water and calorimeter is equal to the heat lost by the quartz sample. Therefore:

m_quartz × c_quartz × ΔT_quartz = (m_water + m_calorimeter) × c_water × ΔT_water

Substituting the given values, we get:

(0.0548 g) × (0.730 J/g·°C) × (T_i - 26.4 °C) = (200.0 g + 100.0 g) × (4.184 J/g·°C) × (26.4 °C - 23.0 °C)

Simplifying and solving for Ti, we get:

Ti = 16.3 °C (rounded to 2 significant digits)

Therefore, the initial temperature of the quartz sample was 16.3 °C.

User Randall Wang
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