Question

A ball of copper weighing 400 gram is transferred from a furnace to a copper calorimeter of mass 300 gram containing 1 kg of water and 20 degree. the temperatures of water rises to 50 degree what is the original temperature of the ball​

Answer 1

Specific heat of water = 4200Jkg-1 K-1

Specific heat of copper= 400 Jkg-1 k-1

Mass of ball= 400gm= 0.4kg

 

According to principle of calorimeter

Mass of calorimeter + water = heat lost by the ball

mcSc dθ+mwSw dθ=Ms dθ

0.3*400* (50-20)+ 1* 4200 (50-20) =0.4* 400*(θ -50)

∴ θ= 860° C

Answer 2

Answer:

The initial temperature of copper ball is 882.26°C

Explanation:

Given that,

Mass of copper ball=400gm=0.4kg

Mass of copper calorimeter=300gm=0.3kg

Mass of water contained in calorimeter=1kg

Initial Temperature of water=20°C=293K

Final Temperature of water=50°C=323K

Let the temperature of the copper ball be T

According to Thermal equilibrium,

Heat lost by copper ball=Heat gained by Calorimeter+Heat gained by water

$= > m _{cu}s _{cu}(T - 323) = m _{w}s _{w}(323 - 293) + m_{cal}s _{cal}(323 - 293)$

The specific heat of Copper is

$s _{cu} = 389J/kgK$

The specific heat of water is

$s _{w} = 4200J/kgK$

Substituting these values in above relation,we get

$0.4 \times 389(T - 323) = 1 \times 4200 \times 30 + 0.3 \times 389 \times 30 \\ = > T = 1155.26K = (1155.26 - 273) {}^{0}C \\ = 882.26 {}^{0} C$

Therefore,the initial temperature of copper ball is 882.26°C

#SPJ3