Question

An unknown silver/gray (92.9-g) piece of metal at 178.0°C is quickly placed to 75.0 mL of water initially at 24.0°C. After a few minutes, both the metal and the water have reached 29.7°C. Calculate the specific heat and the identity of the metal

Answer 1

$TQ(H2O)=c(H2O)×ρ(H2O)×V(H2O)×ΔT$

Q(H2O) = 1795.5 J

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

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

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