For each pair of substances circle the substance with more t

For each pair of substances, circle the substance with more thermal energy and explain why. 500 g of water at 35degree C vs. 500 g of water at 85degree C 500 g of water at 35degree C vs. 250 g of water at 35degree C For each pair of substances, circle the substance that will take more energy to heat from 20degree C to 100degree C and explain why. 30 g of water vs. 300 g of water 500 g of iron vs. 500 g of silver Explain why a can of soda from the refrigerator feels cold in your hand. How much heat must be absorbed by a 500 g pot of water in order to raise the temperature from 20degree C to 30degree C? State the three laws of thermodynamics. What is entropy? Convert 125 degrees Fahrenheit to degrees Celsius and Kelvin. Convert 57 Kelvin to Celsius and Fahrenheit. How much heat must be added to 1.0 kg of ice at 0 degrees Celsius to convert it to water at 0 degrees Celsius? How much heat must be added to 1.0 kg of water at 0 degrees Celsius to convert it into 1.0 kg of water at 100 degrees Celsius? How much heat must be added to 1.0 kg of water at 100 degrees Celsius to convert it to stream at 100 degrees Celsius? How much heat must be added to 1.0 kg of steam at 100 degrees Celsius to convert it to steam at 110 degrees Celsius?

Solution

(6).F= 125 ^oF

Relation between Fahrenheit scale temprature (F) and celcius scale temprature (C)is

(F-32)/180 = C/100

              C = (100/180)(F-32)

                 =(5/9)(125-32)

                 = 51.66 ^o C

                 = 51.66 +273 = 324.66 K

(b).K = 57 K

    In celsius sclae C = (57-273) ^oC

       = -216 ^o C

Relation between Fahrenheit scale temprature (F) and celcius scale temprature (C)is

(F-32)/180 = C/100

       F-32 = (9/5)C

              = (9/5)(-216)

              = - 388.8

          F = -356.8 ^o F

(7).Mass m = 1 kg

Latent heat of fusion of ice L = 333.55 kJ/kg

                                          = 333.55 x10 ³ J / kg

Required heat Q = mL

                         = 333.55 x10 ³ J

(8).Mass m = 1 kg

Specific heat of water C = 4186 J / kg ^o C

Temprature difference dt = 100 ^o C - 0^oC = 100 ^o C

Heat required Q \' = mC dt

                         = 1 kg x 4186 x 100

                         = 4.186 x10 ⁵ J

(9).Latent heat of vaporization L \' = 22.6 x 10⁵ J/kg

Required heat = mL \' = 1 kg x 22.6 x 10⁵ J/kg

                      = 22.6 x 10⁵ J