Question

a calorimeter contains 100g of ice at -20°c.x g of water at 80°c is added to the calorimeter such that the final contents of the calorimeter are in thermal equilibrium with 40°cercury.the value of x is
A 325
B 224
C 60
D 100

please solve this...​

Answer 1

Answer:

I think option C is right answer

if it helps you please mark as brainlest

Answer 2

Given:
Mass of ice(m₁)= 100g

Temperature of ice(t₁)= -20°C

Specific heat capacity of ice(S₁)= 0.49cal/g°C

Mass of water(m₂)=  $xg$

Temperature of water(t₂)= 80°C

Specific heat capacity of water(S₂)= 1cal/g°C

Final temperature of mixture(t)= 40°C

To  find the mass of water

Solution:

Using law of caloriemetry

Heat gained=Heat lost

$m_{1} S_{1} (t_{1}-t) =m_{2}S_{2} (t-t_{2} )$

100×0.49×(-20-40)=$x$×1(40-80)

⇒49×(-60)=$x$×(-40)

⇒ -2940= -40$x$

⇒$x$ =73.5g

Hence, the value of$x$ is 73.5g.