Use the data in the table to answer the following questions. Calculate the energ

Question

Use the data in the table to answer the following questions. Calculate the energy required to heat a beaker of water at 18 degree C to boiling. The mass of the water is 70.0 g. A water heater warms 35-L (35 kg) of water from a temperature of 22.7 degree C to a temperature of 83.7 degree C. Determine the amount of energy (in joules) required. Determine the temperature change that will occur when 250-J of energy is applied to 20. g of gold. When 895-J of heat is applied to a sample of iron metal the temperature increases by 55.0 degree C. Determine the mass of the metal sample. A silver ring has a mass of 138.45 g. How many calories of heat are required to increase the temperature from 11.8 degree C to 162.5 degree C? What is the mass of copper that increases its temperature by 285 degree C when 186.000 J of energy is applied? How much energy (in kJ) is lost by a 348-kg iron statue that goes from a temperature of 299 K to a temperature of 280 K? A 9.84 oz ingot of unknow

Explanation / Answer

Ans. q = m x s x dT            - equation 1   

Where,

q = heat gained (or lost)

m = mass of water in gram

s = specific heat, [of water = 4.184 J g-10C-1]

dT = change in temperature = T1- T1

#1. dT = 1000C – 180C = 820C

m = 70.0 g

Putting the values in equation -1

            q = 70.0 g x (4.184 J g-10C-1) x 820C = 24016.16 J = 24.016 kJ

#2. dT = 83.70C – 22.70C = 610C

m = 35000 g                                     ; [1 kg = 1000 g]

Putting the values in equation -1

            q = 35000 g x (4.184 J g-10C-1) x 610C = 8932840 J = 8932.84 kJ

#3. Putting the values in equation -1

            250 J = 20 g x (0.129 J g-10C-1) x dT

            Or, dT = 250 J / (2.58 J 0C-1) = 96.900C

#4. Putting the values in equation -1

            895 J = m x (0.45 J g-10C-1) x 55.00C

            Or, m = 895 J / (24.75 J g-1) = 36.16 g

#5. dT = 162.50C – 11.80C = 150.70C

Putting the values in equation -1

            q = 138.45 g x (0.24 J g-10C-1) x 150.70C = 5006.73624 J = 1196.64 cal

#6. Putting the values in equation -1

            186000 J = m x (0.387 J g-10C-1) x 2850C

            Or, m = 186000 J / (110.295 J g-1) = 1686.39 g

#7. dT = 280 K – 299 K = -19K = -190C

Note: difference by 1 unit in 0C is equivalent to that of difference by 1 unit in kelvin.

Putting the values in equation -1

            q = 348000 g x (0.45 J g-10C-1) x (-190C) = - 2975400 J = - 2975.4 kJ

Note: the -ve sign of (- 2975.4 kJ) simply indicates that heat is being released during cooling.

#8. m = 9.84 oz = 9.84 x (28.3495 g) = 278.9593 g

191.20F = 84.440C

73.20F = 22.890C

dT = 84.440C - 22.890C = 61.550C

q = 3.91 kcal = 16.35944 kJ = 16359.44 J

Putting the values in equation 1-

            16359.44 J = 278.9593 g x C x 61.550C

            Or, C = 16359.44 J / (17169.944915 g 0C) = 0.953J J g-1 0C-1

Thus, specific heat of the substance = 0.953J J g-1 0C-1

Note: Please refer your reference book for the substance with specific heat nearly equal to 0.953J J g-1 0C-1.

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