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Chapter 10: Problem 30

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 metal rod of Young's modulus \(Y\) and coefficient of thermal expansion \(a\) is held at its two ends such that its length remains invariant. If its temperature is raised by \(t^{\circ} \mathrm{C}\), then the linear stress developed in it is [2011] (A) \(\frac{\alpha t}{Y}\) (B) \(Y \alpha t\) (C) \(\frac{Y}{\alpha t}\) (D) \(\frac{1}{Y \alpha t}\)

The average translational kinetic energy of 1 mole of \(O_{2}\) molecules (molar mass \(=32\) ) at a particular temperature is \(0.048 \mathrm{eV}\). The internal energy of 1 mole of \(N_{2}\) molecules (molar mass \(=28\) ) in \(\mathrm{eV}\) at same temperature is (A) \(0.048\) (B) \(0.003\) (C) \(0.0288\) (D) \(0.080\)

An aluminium sphere of \(20 \mathrm{~cm}\) diameter is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). Its volume changes by (given that the coefficient of linear expansion for aluminium \(\left.\alpha_{A l}=23 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)\) (A) \(28.9 \mathrm{cc}\) (B) \(2.89 \mathrm{cc}\) (C) \(9.28 \mathrm{cc}\) (D) \(49.8 \mathrm{cc}\)

The radius of a metal sphere at room temperature \(T\) is \(R\) and the coefficient of linear expansion of the metal is \(\alpha .\) The sphere heated a little by a temperature \(\Delta T\) so that its new temperature is \(T+\Delta T\). The increase in the volume of the sphere is approximately. (A) \(2 \pi R \alpha \Delta T\) (B) \(\pi R^{2} \alpha \Delta T\) (C) \(4 \pi R^{3} \alpha \Delta T / 3\) (D) \(4 \pi R^{3} \alpha \Delta T\)

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