Question

1. Consider the following data for aluminum, Cu, and an unknown metal z cu metal Z 10. Metal Mass (8) Specific heat (3/8°C) Initial temperature (°C) Final temperature (Al/cu) Final temperature (Al/Z) 10.0 0.900 40.00 51.25 48.50 30.0 0.385 60.00 51.25 55.00 48.50 a) When Al and Cu are placed in contact with each other, the final temperature of the metals is 51.2°C How much heat is exchanged between the two metals? b) When Al and the unknown metal Z are in contact with each other, the final temperature is 48.5°C. Without doing the calculation, what can you predict about the specific heat capacity of the metal Z? Explain briefly but clearly. c) Now calculate the specific heat capacity of the unknown metal, Z. (include units!!) Specific heat 2 (c) = 2 M.AT

Answer 1

1.

a)

Mass of Al

Specific heat of Al $C=0.900 \ \frac{J}{g \cdot ^oC}$

Temperature change for Al $\Delta T= 51.25 \ ^oC-40.00 \ ^oC=11.25 \ ^oC$

Heat gained by Al

д%3D тхСX A7

$q =10.0 \ g \times 0.900 \ \frac{J}{g \cdot ^oC} \times 11.25 \ ^oC$

Hence, the heat exchanged between Al and Cu is 101.25 J.

b)

Mass of Cu is 30.0 g. It is higher than the mass of Z which is 10.0 g.

The temperature change for Cu is higher than the temperature change for Z.

Hence, the specific heat for Z is predicted to be higher than that of Cu.

The temperature change for Al is higher than the temperature change for Z.

Hence, the specific heat for Z is predicted to be higher than that of Al.

c)

Mass of Al

Specific heat of Al $C=0.900 \ \frac{J}{g \cdot ^oC}$

Temperature change for Al $\Delta T= 48.50 \ ^oC-40.00 \ ^oC=8.50 \ ^oC$

Heat gained by Al

д%3D тхСX A7

$q =10.0 \ g \times 0.900 \ \frac{J}{g \cdot ^oC} \times 8.50 \ ^oC$

Hence, the heat exchanged between Al and Z is 76.5 J.

Mass of Z

Specific heat of Z is

Temperature change for Z $\Delta T= 48.50 \ ^oC-55.00 \ ^oC=-6.50 \ ^oC$

Heat lost by Z

$q =-m \times C \times \Delta T$

$76.5 \ J =-10.0 \ g \times C \times (-6.50 \ ^oC)$

$C =1.18 \ \frac{J}{g \cdot ^oC}$

Hence, the specific heat of Z is $1.18 \ \frac{J}{g \cdot ^oC}$ .