1) What is the speed of sound along Roxas Boulevard in Manila in the summer when the temperature is 32°C? Compare this with the speed of sound in December when the temperature is 19°C. 2.1f you are 600m from a batter when you see him hit a ball, how long will you wait to hear the sound of the bat hitting the ball {at 25°C]? 3.A.1ightning flash was seen 2 seconds before the clap of thunder. If the temperature was 25°C, how far away did the lightning occur? 4.An explosion occurs at a distance of 6.0 km from a person. How long after the explosion does the person hear it? Assume the temperature is 14°C. 5.Eight seconds after a sound was emitted by a sonic altimeter, the echo is heard. How high is the airplane if the temperature is 10°C? 6.A.ship sails away from an iceberg. Determine the temperature of the air if the distance of the ship from the iceberg is 846.3m after 2.65. VU°C= 331.5

1. v2?=≈349.97m/s when T2?=32oC and  the speed of sound in the summer at T2?=32oC is 1.02 times greater than the speed of sound in December when T1?=19oC.

 

2. t=1.73s

 

3. s≈691.86m

 

4. t≈17.67s

 

5. s≈2697.96m

 

6. T≈263.97K≈−9.18oC

Step-by-step explanation

NOTE:

•  speed of sound(v):    v=MγRT??, Equation (1)

            where:

             γ=adiabaticconstant≈1.4 (for air)

             R=gas constant=8.314 J/mol-K

             T= absolute temperature

             M=molecular mass of gas=29 g/mol

 

• speed(v)=speed,v=time,tdistance,s? , Equation (2)

 

1. Given:     T1?=19oC     v1?=?

                   T2?=32oC     v2?=?

 

Solution: 

STEP 1: Solving for v2?, we use Equation (1) at T2?=32oC=305.15K. So,

v2?=MγRT2???=(29molg?)(1000g1kg?)(1.4)(8.314mol−KJ?)(305.15K)??≈349.97m/s

 

STEP 2:  Equation (1) can also be expressed as: v2=MγRT?

STEP 3: Assume mass of air is constant as well as R and γ. 

Equating the equation to MγR?=constant, we have:  Tv2?=MγRT?

Hence, 

            (MγRT?)1?=(MγRT?)2?

                     T1?v12??=T2?v22??

STEP 4: Substitute the values of T1?=19oC and T2?=32oC.

292.15v12??=305.15v22??

 

STEP 5: Solve for v2?,

1.6842v12?=v22?

v2?=1.02v1?

Therefore, the speed of sound in the summer at T2?=32oC is 1.02 times greater than the speed of sound in December when T1?=19oC.

 

2. 

STEP 1: 

Calculating for the speed of sound (v) at T=25oC=298.15K with the use of Equation (1), we get: v=345.93m/s

 

STEP 2: Solving for t using Equation 2:

             speed,v=time,tdistance,s?

       345.93m/s=time,t600m?

                         t=1.73s

 

3.

STEP 1: 

Calculating for the speed of sound (v) at T=25oC=298.15K with the use of Equation (1), we get: v=345.93m/s

 

STEP 2: Solving for s using Equation 2:

             speed,v=time,tdistance,s?

       345.93m/s=2ss?

                         s≈691.86m

 

4.

STEP 1: 

Calculating for the speed of sound (v) at T=14oC=297.15K with the use of Equation (1), we get: v=339.49m/s

 

STEP 2: Solving for t using Equation 2:

             speed,v=time,tdistance,s?

       339.49m/s=t6km(1km1000m?)?

                         t≈17.67s

 

5.

STEP 1: 

Calculating for the speed of sound (v) at T=10oC=283.15K with the use of Equation (1), we get: v=337.12m/s

 

STEP 2: Solving for s using Equation 2:

             speed,v=time,tdistance,s?

       337.12m/s=8ss?

                         s≈2696.96m

 

6.

STEP 1: 

Solving for v using Equation 2:

             speed,v=time,tdistance,s?

            speed,v=2.6s846.3m?

                         v=≈325.5m/s

STEP 2: 

Calculate for T,

v=MγRT??

325.5sm?=(29molg?)(1000g1kg?)1.4(8.314mol−KJ?)T??

T≈263.97K≈−9.18oC