Question

A 50.0 g sample of an unknown substance absorbed 1.64 kJ of energy as it changed from a temperature of 36°C to 98°C. What is the specific heat of the unknown substance in J/g°C?

Answer 1

`"0.53 J/g°C"`

Explanation:

Specific heat of a substance is given by the formula

`"S" = "Q"/"mΔT"`

`"S" = (1.64 × 10^3\ "J")/"50.0 g × (98 – 36)°C" = "0.53 J/g°C"`