A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal? 3.433 0.171 1.717 0.343

The specific heat of metal is c = 3.433 J/g*⁰C.

Given

mass of metal = 68.6 g

t metal = 100 °C

mass water = 84 g

t water = 20 °C

final temperature = 52.1  °C

Required

The specific heat

Solution

Heat can be formulated :

Q = m.c.Δt

Q absorbed by water = Q released by metal

84 x 4.184 x (52.1-20)=68.6 x c x (100-52.1)

11281.738=3285.94 x c

c = 3.433 J/g*⁰C.

i do not get it   at all