Question

1. light with a frequency of 7.26×10¹⁴ hz lies in the violet region of the visible spectrum. the wavelength of this frequency of light is (a) 435×10⁻⁸ m (b) 456×10⁻¹⁰ m (c) 1000 km (d) none of the above 26. the total amount of heat needed to change 40.0 g of 0°c ice to steam at 100°c is (a) 333×10⁸ j (b) 54665 j (c) 5600 j (d) 7689 j (e) none of the above assume the following c = 1 cal/g°c

Explanation:

Question 25

Step1: Recall the formula for the speed of light

The speed of light \(c = \lambda f\), where \(c = 3\times10^{8}\space m/s\), \(\lambda\) is the wavelength and \(f\) is the frequency. We need to solve for \(\lambda\), so \(\lambda=\frac{c}{f}\)

Step2: Substitute the given values

Given \(f = 7.26\times 10^{14}\space Hz\) and \(c = 3\times 10^{8}\space m/s\). Then \(\lambda=\frac{3\times 10^{8}}{7.26\times 10^{14}}\)

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Now check the options:

• Option (a): \(435\times 10^{-8}\space m=4350\space nm\)
• Option (b): \(456\times 10^{-10}\space m = 45.6\space nm\)
• Option (c): \(1000\space km = 10^{6}\space m\)

Step1: Calculate the heat for melting ice

The heat of fusion \(L_f=334\space J/g\). For \(m = 40\space g\) of ice, \(Q_1=mL_f=40\times334 = 13360\space J\)

Step2: Calculate the heat for heating water from \(0^{\circ}C\) to \(100^{\circ}C\)

The specific heat of water \(c = 4.186\space J/g^{\circ}C\). \(Q_2=mc\Delta T\), \(\Delta T=(100 - 0)^{\circ}C\), \(m = 40\space g\). So \(Q_2=40\times4.186\times100=16744\space J\)

Step3: Calculate the heat for vaporizing water

The heat of vaporization \(L_v = 2260\space J/g\). For \(m = 40\space g\) of water, \(Q_3=mL_v=40\times2260=90400\space J\)

Step4: Calculate the total heat

\(Q_{total}=Q_1 + Q_2+Q_3\)

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Now check the options:

• Option (a): \(333\times 10^{8}\space J\)
• Option (b): \(54665\space J\)
• Option (c): \(5600\space J\)
• Option (d): \(7689\space J\)

Answer:

d. none of the above

Question 26