# Chemistry Question - Calorimetry?

I don't understand how to solve this problem:

When a 12.8g sample of KCl dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0°C to 21.6°C. Calculate the ∆H for the process:

KCl → K+ + Cl-

(The + and - in the K and Cl are superscripts.)

The equation to use is

q = m(SpHt)(delta T)

q is the amount of heat gained or lost = delta H for this process

m is the mass of the solution = 87.8g (12.8g + 75.0g)

delta T is the change in temperature during the process 31.0 oC - 21.6 oC = 9.4 oC

SpHt is the specific heat of the solution = 4.184 J / g oC

NOTE: Since the specific heat of the resulting KCl solution is not given, it will be assumed that it is the same as the specific heat of water.

q = (87.8g)(4.184J / g oC)(9.4 oC)

NOTE: Since the temperature is decreasing, it is an endothermic reaction so the values of q and delta H will be positive values.

q = +345J = delta H

q = mCdeltaT

C = specific heat of water - check your units, there is a C for oC and a C for Kelvin

delta T = change in temp - if you use the C that has oC then just subtract your temps! if Kelvin - convert!

m = mass

q at a constant pressure - which this is = delta H

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When a 12.8g sample of KCl dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0°C to 21.6°C. Calculate the ∆H for the process:

KCl → K+ + Cl-

(The + and - in the K and Cl are superscripts.)

**Answer:**The equation to use is

q = m(SpHt)(delta T)

q is the amount of heat gained or lost = delta H for this process

m is the mass of the solution = 87.8g (12.8g + 75.0g)

delta T is the change in temperature during the process 31.0 oC - 21.6 oC = 9.4 oC

SpHt is the specific heat of the solution = 4.184 J / g oC

NOTE: Since the specific heat of the resulting KCl solution is not given, it will be assumed that it is the same as the specific heat of water.

q = (87.8g)(4.184J / g oC)(9.4 oC)

NOTE: Since the temperature is decreasing, it is an endothermic reaction so the values of q and delta H will be positive values.

q = +345J = delta H

q = mCdeltaT

C = specific heat of water - check your units, there is a C for oC and a C for Kelvin

delta T = change in temp - if you use the C that has oC then just subtract your temps! if Kelvin - convert!

m = mass

q at a constant pressure - which this is = delta H

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