Question

A student heats 66.90 grams of silver to 98.61 °C and then drops it into a cup containing 81.04 grams of water at 24.12 °C. She measures the final temperature to be 27.32 °C. The heat capacity of the calorimeter was determined in a separate experiment to be 1.87 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of silver.

Answer 1

metal

mass of metal = 66.90 g ,

T1 = 98.61 oC

T2 = 27.32 oC

dT =T2 - T1 = 27.32 -98.61 = -71.29 oC

q = m Cp dT

= 66.90 x Cp x (-71.29 )

water :

mass of water = 81.04 g

T1 = 24.12

T2 = 27.32

dT = 3.2

q = m Cp dT

= 81.04 x 4.184 x 3.2

= 1085 J

heat loss by the metal = heat gain by the water

- 66.90 x Cp x (-71.29 ) = 1085

Cp = 0.228 J / g oC

specific heat of metal =  0.228 J / g ^oC