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Chapter 14: Problem 99

The inner vessel of a calorimeter contains \(2.50 \times 10^{2} \mathrm{g}\) of tetrachloromethane, \(\mathrm{CCl}_{4},\) at \(40.00^{\circ} \mathrm{C} .\) The vessel is surrounded by \(2.00 \mathrm{kg}\) of water at $18.00^{\circ} \mathrm{C} .\( After a time, the \)\mathrm{CCl}_{4}$ and the water reach the equilibrium temperature of \(18.54^{\circ} \mathrm{C} .\) What is the specific heat of \(\mathrm{CCl}_{4} ?\)

Short Answer

Expert verified
Answer: The specific heat capacity of tetrachloromethane (CCl4) is approximately \(0.91 \mathrm{J}/(\mathrm{g} \cdot \mathrm{°C})\).

Step by step solution

01

Understand the Concept of Heat Exchange

In this exercise, we have a system with two substances: CCl4 and water. When they reach equilibrium, the heat gained by one substance is equal to the heat lost by the other substance, assuming no heat loss to the surroundings. This is the basic principle of calorimetry, which we will use to solve this problem.
02

Write the Heat Exchange Formula

The formula for heat transfer (Q) is given by the equation: Q = mcΔT where m is the mass of the substance, c is its specific heat, and ΔT is the change in temperature. In this case, we will have two separate equations - one for CCl4 and one for water - and we will set them equal to one another.
03

Write the Equation for CCl4

For CCl4, we have: Q_CCl4 = m_CCl4 * c_CCl4 * ΔT_CCl4 Here, m_CCl4 = \(2.50 \times 10^{2} \mathrm{g}\), c_CCl4 is the specific heat that we need to find, and ΔT_CCl4 = (T_final - T_initial) = \(18.54^{\circ} \mathrm{C} - 40.00^{\circ} \mathrm{C}=-21.46^{\circ} \mathrm{C}\)
04

Write the Equation for Water

For water, we have: Q_water = m_water * c_water * ΔT_water Here, m_water = \(2.00 \mathrm{kg}\) = \(2000 \mathrm{g}\), c_water \(= 4.18 \mathrm{J}/(\mathrm{g} \cdot \mathrm{°C})\) (specific heat of water), and ΔT_water = (T_final - T_initial) = \(18.54^{\circ} \mathrm{C} - 18.00^{\circ} \mathrm{C} = 0.54^{\circ} \mathrm{C}\)
05

Set the Equations Equal

Now, we can set the heat exchange equations equal to one another, as we know that the heat lost by CCl4 equals the heat gained by the water: m_CCl4 * c_CCl4 * ΔT_CCl4 = m_water * c_water * ΔT_water
06

Solve for the Specific Heat of CCl4

We can now solve the equation for the specific heat of CCl4: c_CCl4 = (m_water * c_water * ΔT_water) / (m_CCl4 * ΔT_CCl4) Plugging in the values, we get: c_CCl4 = (\(2000 \mathrm{g} * 4.18 \mathrm{J}/(\mathrm{g} \cdot \mathrm{°C}) * 0.54^{\circ} \mathrm{C}\)) / (\(2.50 \times 10^{2} \mathrm{g} * -21.46^{\circ} \mathrm{C}\)) c_CCl4 = \(\approx 0.91 \mathrm{J}/(\mathrm{g} \cdot \mathrm{°C})\)
07

Write the Final Answer

The specific heat capacity of tetrachloromethane (CCl4) is approximately \(0.91 \mathrm{J}/(\mathrm{g} \cdot \mathrm{°C})\).

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Most popular questions from this chapter

While camping, some students decide to make hot chocolate by heating water with a solar heater that focuses sunlight onto a small area. Sunlight falls on their solar heater, of area \(1.5 \mathrm{m}^{2},\) with an intensity of $750 \mathrm{W} / \mathrm{m}^{2}$ How long will it take 1.0 L of water at \(15.0^{\circ} \mathrm{C}\) to rise to a boiling temperature of $100.0^{\circ} \mathrm{C} ?$
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