How many grams of silver are needed to absorb 1.00 kJ of energy changing the temperature from 33.0°C to 41.8°C?

1 Answer
Apr 15, 2017

Answer:

#484g#

Explanation:

To solve this, we use the formula for heat absorption, #Q = mTc#, where
#Q# is the heat energy absorbed, in Joules,
#m# is the mass of the object usually in grams,
#T# is the temperature change of the object, in Celcius or Kelvin, and
#c# is the specific heat capacity of the material, in joules per gram °C. This is the amount of energy it takes for one gram of this substance to heat up by 1K.

We then rearrange this equation making #m# the subject:
#m = Q/(Tc)#

The question gives us the values #Q = 1kJ = 1000J#, and we can calculate #T# by using

#T = "Final Temperature" - "Initial Temperature" = 41.8"°C" - 33.0"°C"#
#T = 8.80"°C = 8.80K"#

Now the only remaining value we need is #c#. Since this depends on the material involved, we have to look up the value for silver.
I found 3 sources: hyperphysics, periodictable.com, and ptable.com.

These put the value at #0.233Jg^-1K^-1#, #0.235Jg^-1K^-1#, and #0.235Jg^-1K^-1# respectively. I will use #0.235Jg^-1K^-1# here.

Now we substitute into the equation:

#m = Q/(Tc) = (1000J)/(8.80K xx 0.235Jg^-1K^-1) = 484g " (3 s.f.)"#

If you have been given a value for the specific heat capacity of silver which is not #0.235Jg^-1K^-1#, then your final answer may be slightly different.