Question

A 92.9 piece of a silvery grey metal is heated to 178 degree celsius, and then quickly transferred into 75.0 mL of water initially at 24.0 degree celsius After 5 minutes both the metal and the water have reached the same temperature of 29.7 degrees celsius

what is the specific heat of the metal and identify the metal

Answer 1

Solution.

$Q(H2O) = c(H2O) \times \rho(H2O) \times V(H2O) \times \Delta T$

Q(H2O) = 1795.5 J

$c(Me) = \frac{Q(H2O)}{m(Me) \times \Delta T1}$

$c(Me) = \frac{1795.5}{0.0929 \times (178-29.7)} = 130.33 \frac{J}{kg \times ^oC}$

Therefore, it is lead.

Answer:

$c(Me) = 130.33 \frac{J}{kg \times ^oC}$

Lead.