Answer to Question #127623 in General Chemistry for Silvia Attia

Question

In the laboratory a student finds that it takes 25.0 Joules to increase the temperature of  10.1 grams of  liquidmercuryfrom  20.2 to 39.7 degrees Celsius.

The specific heat of mercurycalculated from her data is  J/g°C.

Answer 1

specific heat = (amount of heat in J)/(mass of substance in g)(temperature change in °C)

temperature change Δt = 39.7 °C – 20.2 °C = 19.5 °C

specific heat = 25.0 J / (10.1 g × 19.5 °C) = 0.127 J/g°C

Answer: 0.127 J/g°C