Q1.
m = 19.29 g of metal
Tm = 99.7°C
V = 50 mL o fwater, T = 23.4°C
final Mx T = 26.4°C
find Cp metal.
The energy balance:
Qgain = -Qloss
Qwater= -Qmetal
Qwater = mwater * Cp water * (Tf-Twater)
Qmetal = m metal * Cp metal * (Tf-Tmetal)
then
mwater * Cp water * (Tf-Twater)= -m metal * Cp metal * (Tf-Tmetal)
substitute knonw data
mwater = Dwater*Vwater = 1 g/mL *50 mL = 50 g
50 *4.184 * (26.4-23.4)= -19.29 * Cp metal * (26.4-99.7)
Cp metal = 50 * 4.184 * (26.4-23.4) / (-19.29 * (26.4-99.7) )
Cp metal = 0.4438 J/gC for this metla
Q2.
possible losses:
- Ambient
- Vaporization of water
- Sound/Work done by systems
- Loss due to container low/poor adiabatic status
Q3
Since Q is lost, then The temperature of the mtal will be lower, than the theoretical expected
if this is true, recall that
C metal = Q/(mass*dT)
if dT is lower, then, Cp metal is higher (incorrectly)
Q4
We can calculate the specific heat of the calorimeter in another experiment, this will be much precise
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