/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Two liquids \(A\) and \(B\) are ... [FREE SOLUTION] | 91影视

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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

An inflated rubber balloon contains 1 mole of an ideal gas, has a pressure \(p\), volume \(V\), and temperature \(T\). If the temperature rises to \(1.1 T\), and the volume is increased to \(1.05 \mathrm{~V}\), the final pressure will be (A) \(1.1 p\) (B) \(p\) (C) less than \(p\) (D) between \(p\) and \(1.1\)

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