1) A 50.0 gram sample of water is heated from 21.5 degrees C to 26.21 degrees C.

Question

1) A 50.0 gram sample of water is heated from 21.5 degrees C to 26.21 degrees C. How many Joules of heat were added to this solution?

2) A 15.25 gram sample of copper initially at 92 degrees is added to a container containing water. The final temperature of the metal is 26.21 degrees C. What is the total amount of energy in Joules added to the water? What was the energy lost by the metal?

3) Mixing 25.0 mL of 1.2 M HClO4 and 25.0 mL of 1.1 M NaOH were mixed. The temperature of the initial solution was 22.4 degrees C. Assuming a Heat of Neutralization of 55.8 KJ/mol, what would the final temperature be if the specific heat for this solution is 4.032 J/g o ?

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Explanation / Answer

For an approximate calculation:

specific heat capacity for water = 4.18 J/(g*degC)
(how much energy is required per gram per change in degrees C)

mass = 50g
Change in temperature = 26.21-21.5 = 4.71 degC

energy required = mass * change in temperature * specific heat capacity
= 50g * 4.71 degC * 4.18 J/(g*degC)

=984.39J

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