Question

Data Tables 1. Experiment 2A Trial #1 Trial #2 15.61g 100.0°C 15.61g 100.0°C 99.74g 100.00 g Metal Used= Copper Mass of metal, m (g) Initial temperature of metal (Tı) Mass of water, m (g) Initial temperature of water (T1) Final temperature of the system (T2) AT (Water) AT (Metal) 26.5°C 26.5°C 27.1°C 27.9°C 0.6°C 1.4°C 72.9°C 72.1°C Q water J J Cp for the metal J/g.°C J/g.°C

Find Q Water for Trial #1 and Trial #2
Find Cp for the metal for trial #1 & #2

Answer 1

Dear student the calorimetry formula Q water as shown below,

$Q_{water}= mC\Delta T$

here for water specific heat C= 4.184 J/kg⋅°C

Now, for Trial 1 :-

$Q_{water}= mC\Delta T$

m = 100 g, C= 4.184 J/kg⋅°C,  ΔT = 0.6°C

$Q_{water}= 100\times 4.184\times 0.6$

Qwater 251.04J

Simillary for Trial 2 :-

$Q_{water}= mC\Delta T$

m = 99.74 g, C= 4.184 J/kg⋅°C,  ΔT = 1.4°C

Qwater 99.74 x 4.184 x 1.4

Qwater 584.24J

Now, we calculate Cp of metal for trial 1 and 2 :-

As we know the Qwater so we can compare with Qmetal also hence we get the Cp of metal from that equation,

Q_water = Q_metal

$m\times Cp_{water}\times \Delta T = m\times Cp_{metal}\times \Delta T$

For Trial 1 :-

$100\times 4.184\times 0.6 = 15.61 \times Cp\times 72.9$

$Cp=\frac{100\times 4.184\times 0.6}{15.61 \times72.9}$

$Cp=\frac{251.04}{1137.97}$

Cp = 0.22 J/kg⋅°C

For Trial 2 :-

$99.74\times 4.184\times 1.4 = 15.61 \times Cp\times 72.1$

$Cp=\frac{99.74\times 4.184\times 1.4}{15.61 \times 72.1}$

$Cp=\frac{584.24}{1125.48}$

Cp = 0.591 J/kg⋅°C

THANK YOU FOR VIEWING MY ANSWER IF YOU HAVE ANY QUERY/DOUBT/OPINION ABOUT ABOVE ANSWER PLEASE COMMENT BELOW AND RATE MY ANSWER BY THUMBS UP AND DOWN.