How much heat energy is required to raise the temperature of 25.0 g of solid NaC

Question

How much heat energy is required to raise the temperature of 25.0 g of solid NaCl from -11.3 degree to 86.2 degree, given that the specific heat capacity of NaCl is 0.864 J/g degree? A 22.31 -g piece of beryllium at 58.7 degree is placed into 41.33 g of water at 16.1 degree. What will the temperature be after the water and the beryllium have reached thermal equilibrium, assuming that no heat is lost to the surroundings? The specific heat capacity of beryllium is 1.82 J/g degree, and that of water is 4.18 J/g degree. A chemist puts 50.0 g of aluminum at 68.0 degree and an unknown mass of titanium at 75.0 degree into 125.0 g of water at 6.3 degree. When the mixture reaches thermal equilibrium, the temperature is 14.7 degree. Using this information, and assuming that no heat is lost to the surroundings, calculate the mass of the piece of titanium. The relevant specific heat capacities are: 0.902 J/g degree (Al), 0.522 J/g degree (Ti), and 4.18 J/g degree (water). How much heat is required to convert 50.0 g of water at 25.0 degree into steam at 150.0 degree, given the following information: specific heat capacity of water = 4.18 J/g degree, specific heat capacity of steam = 2.04 J/e degree, boiling point of water = 100.0 degree, heat of vaporization of water is 2257 J/g. A 43.75-g piece of gallium metal at 5.2 degree is placed into 61.99 g of water at 41.6 degree. What will be the temperature when this mixture reaches thermal equilibrium, assuming that no heat is lost to the surroundings? Use the following information: specific heat capacity of gallium = 0.371 J/g degree, specific heat capacity of water = 4.18 J/g degree, melting point of gallium = 29.8 degree, heat of fusion of gallium = 5.56 kJ/mol.

Explanation / Answer

1) 25*0.864*(86.2-(-11.3)) = 2106 J

2) 22.31*1.82(58.7-T)= 41.33*4.18*(T-16.1)

T= 24.2 Centigrade

3)50*0.902*(68.0-14.7)+x*0.522*(75-14.7) = 125*4.18*(14.7-6.3)

mass of Al = 63.06 grams

4) heat required = mass of water*specific heat of water* ( 100- initial temp) + mass of water * heat of vapourisation + mass of steam*specificheat of steam*(final temp- 100)

heat required= 50*4.18*(100-25) + 50*2257+50*2.04*(150-100) = 133.625 KJ

5)thermal equilibrium temp = T

heat lost by water = heat gained by gallium

61.99*4.18*(41.6-T)= 43.75*0.371*(29.8-5.2) + (x grams /69.723)*5560

T = 29.8o C and only 33.335 grams of gallium is melted to liquid form and remaining gallium( 10.41 g) is solid at 29.8 centrigrade

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