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Chapter 10: Problem 38

Two liquids \(A\) and \(B\) are at \(32^{\circ} \mathrm{C}\) and \(24^{\circ} \mathrm{C}\). When mixed in equal masses, the temperature of the mixture is found to be \(28^{\circ} \mathrm{C}\). Their specific heats are in the ratio of (A) \(3: 2\) (B) \(2: 3\) (C) \(1: 1\) (D) \(4: 3\)

Short Answer

Expert verified
The ratio of specific heats of the two liquids A and B can be found by dividing the temperature difference of the mixture and liquid B by the temperature difference of liquid A and the mixture. Plugging in the given values, we get \(\frac{28 - 24}{32 - 28} = \frac{c_A}{c_B}\), which simplifies to \(\frac{4}{4} = \frac{c_A}{c_B}\) or \(\frac{c_A}{c_B} = 1\). Therefore, the specific heats are in the ratio 1:1, which corresponds to option (C).

Step by step solution

01

List all given information

We are given: 1. Temperature of liquid A: \(T_A = 32^{\circ} \mathrm{C}\) 2. Temperature of liquid B: \(T_B = 24^{\circ} \mathrm{C}\) 3. Temperature of mixture when A and B are mixed in equal masses: \(T_{mixture} = 28^{\circ} \mathrm{C}\) 4. The specific heats of A and B are unknown and denoted by \(c_A\) and \(c_B\) respectively.
02

Use the formula for heat exchange

We know that the heat gained by one substance must equal the heat lost by the other substance when mixed. In this case, heat gained by liquid B = heat lost by liquid A. We'll use the formula for heat exchange: \(m_Bc_BT_{mixture} - m_Bc_BT_B = m_Ac_AT_A - m_Ac_AT_{mixture}\) Since the masses of A and B are equal, we can denote them both as m: \(mc_B(T_{mixture} - T_B) = mc_A(T_A - T_{mixture})\)
03

Solve for the ratio of specific heats

We want to find the ratio of specific heats \(\frac{c_A}{c_B}\). Divide both sides by \(mc_B c_A\): \(\frac{T_{mixture} - T_B}{T_A - T_{mixture}} = \frac{c_A}{c_B}\) Now, plug in the given values for the temperatures: \(\frac{28 - 24}{32 - 28} = \frac{c_A}{c_B}\) \(\frac{4}{4} = \frac{c_A}{c_B}\) \(\frac{c_A}{c_B} = 1\) Thus, the ratio of specific heats is 1:1 which corresponds to option (C).

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Most popular questions from this chapter

A \(2 \mathrm{gm}\) bullet moving with a velocity of \(200 \mathrm{~m} / \mathrm{s}\) is brought to a sudden stoppage by an obstacle. The total heat produced goes to the bullet. If the specific heat of the bullet is \(0.03 \mathrm{cal} / \mathrm{gm}-{ }^{\circ} \mathrm{C}\), the rise in its temperature will be (A) \(158.0^{\circ} \mathrm{C}\) (B) \(15.80^{\circ} \mathrm{C}\) (C) \(1.58^{\circ} \mathrm{C}\) (D) \(0.1580^{\circ} \mathrm{C}\)

At what temperature, the Fahrenheit and the Celsius scales will give numerically equal (but opposite in sign) values? (A) \(-40^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{C}\) (B) \(11.43^{\circ} \mathrm{F}\) and \(-11.43^{\circ} \mathrm{C}\) (C) \(-11.43^{\circ} \mathrm{F}\) and \(+11.43{ }^{\circ} \mathrm{C}\) (D) \(+40^{\circ} \mathrm{F}\) and \(-40^{\circ} \mathrm{C}\)

At what temperature, the Fahrenheit and the Celsius scales will give numerically equal (but opposite in sign) values? (A) \(-40^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{C}\) (B) \(11.43^{\circ} \mathrm{F}\) and \(-11.43^{\circ} \mathrm{C}\) (C) \(-11.43^{\circ} \mathrm{F}\) and \(+11.43^{\circ} \mathrm{C}\) (D) \(+40^{\circ} \mathrm{F}\) and \(-40^{\circ} \mathrm{C}\)

The temperature of a substance increases by \(27^{\circ} \mathrm{C} .\) On the Kelvin scale, this increase is equal to (A) \(300 \mathrm{~K}\) (B) \(2.46 \mathrm{~K}\) (C) \(27 \mathrm{~K}\) (D) \(7 \mathrm{~K}\)

A vessel of volume \(V\) contains a mixture of 1 mole of hydrogen and 1 mole oxygen (both considered as ideal). Let \(f_{1}(v) d v\) denote the fraction of molecules with speed between \(v\) and \((v+d v)\) with \(f_{2}(v) d v\), similarly for oxygen. Then, (A) \(f_{1}(v)+f_{2}(v)=f(v)\) obeys the Maxwell's distribution law. (B) \(f_{1}(v), f_{2}(v)\) will obey the Maxwell's distribution law separately. (C) neither \(f_{1}(v)\) nor \(f_{2}(v)\) will obey Maxwell's distribution law. (D) \(f_{2}(v)\) and \(f_{1}(v)\) will be the same.
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