A 50.0 g silver spoon at 20.0 degrees Celsius is placed in a cup of coffee at 90.0 degrees C. How much heat does the spoon absorb from the coffee to reach a temp. of 89.0 degrees C?
Rememeber this equation. It, or a slight modification of it, will work all of these heat problems.
q = mc(Tf - Ti)
q is the heat lost or gained (depending upon the problem.
m is the mass, usually in grams.
c is the specific heat. If m is in grams then specific heat must be in cal/g*c or joules/g*c depending upon the unit you want for q.
Tf is the final T.
Ti is the initial T.
You will need to look up the specific heat of silver if it isn't given in the problem.
I hope this helps. Post your work if you get stuck.
2 answers
193.2
q = mcTc
m = 50.0 g = 0.050 kg
c = specific heat capacity of silver = 234.3 J kg-1 °C -1
Tc = Change in temperature = 89°C – 20°C = 69 °C
q = 0.050 kg * 234.3 J kg-1 °C -1* 69 °C
Heat gained (q) = 8.1*10^2 J.
m = 50.0 g = 0.050 kg
c = specific heat capacity of silver = 234.3 J kg-1 °C -1
Tc = Change in temperature = 89°C – 20°C = 69 °C
q = 0.050 kg * 234.3 J kg-1 °C -1* 69 °C
Heat gained (q) = 8.1*10^2 J.
0 answers