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Chapter 10: Problem 8

Minimum amount of steam of \(100^{\circ} \mathrm{C}\) required to melt 12 gm ice completely will be (A) \(1.5 \mathrm{gm}\) (B) \(1 \mathrm{gm}\) (C) \(2 \mathrm{gm}\) (D) \(5 \mathrm{gm}\)

Short Answer

Expert verified
The minimum amount of steam at \(100^{\circ} \mathrm{C}\) required to completely melt 12 g of ice is \(1.5 \mathrm{g}\), so answer (A) is correct.

Step by step solution

01

Calculate the Energy Required to melt Ice.

Since the ice is at 0掳C, the energy (\(Q_1\)) required to melt the ice can be described by the equation: \(Q_1=m_1 \cdot L_f\) where \(m_1\) is the mass of the ice (12g) and \(L_f\) is the specific latent heat of fusion for the water (334 J/g). Hence, \(Q_1 = 12 \cdot 334 = 4008 \, J\).
02

Calculate the Energy Released by Steam.

The energy (\(Q_2\)) released by the steam can be calculated using the equation: \(Q_2=m_2 \cdot (C_w \cdot \Delta T + L_v)\), where \(m_2\) is the amount of steam, which is what we want to find, \(C_w\) is the specific heat of water (4.18 J/g掳C), and \(L_v\) is the specific latent heat of vaporization for water (2260 J/g). Because the steam goes from 100掳C to 0掳C and then condenses, \(\Delta T = 100掳C\).
03

Set the two Energy Equal and Solve.

The amount of energy needed to melt the ice is equal to the amount of energy released by the steam, \(Q_1 = Q_2\). Solve this equation for \(m_2\):\n\(4008 = m_2 \cdot (4.18 \cdot 100 + 2260)\) to find \(m_2 = 1.5 \, g\).

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Most popular questions from this chapter

On the Celsius scale, the absolute zero of temperature is at (A) \(0^{\circ} \mathrm{C}\) (B) \(-32^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(-273.15^{\circ} \mathrm{C}\)

A wooden wheel of radius \(R\) is made of two semicircular parts (see Fig. 10.24). The two parts are held together by a ring made of a metal strip of crosssectional area \(S\) and length \(L . L\) is slightly less than \(2 p R\). To fit the ring on the wheel, it is heated so that its temperature rises by \(\Delta T\) and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semi-circular parts together. If the coefficient of linear expansion of the metal is \(a\) and its Young's modulus is \(Y\), then the force that one part of the wheel applies on the other part is (A) \(2 \pi S Y \alpha \Delta T\) (B) \(S Y \alpha \Delta T\) (C) \(\pi S Y \alpha \Delta T\) (D) \(2 S Y \alpha \Delta T\)

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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 \(-40{ }^{\circ} \mathrm{C}\)
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