Question

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 20.72 °C are mixed. The specific heat of water is 4.184 J/gC). Record your answer in scientific notation using 3 significant figures.

Answer 1

Given,

Mass of sample of water(m₁) = 50.0 g

The initial temperature of the sample of water(T_i)₁ = 100 ⁰C

Mass of sample of water(m₂) = 100 g

The initial temperature of the sample of water(T_i)₂ = 20.72 ⁰C

Specific heat of water = 4.184 J/g ^oC

When water at 100 ^oC is added to the water at 20.72 ^oC,

Heat gained by 100 g of water = Heat loss by 50.0 g of water

m₂ x C x T = - (m₁ x C x T)

Here,

T = T_f -T_i

and C is the specific heat of the water.

Neglecting "C" from the above formula, since it is same on both the sides,

m₂ x T = - (m₁ x ​​​​​​​T)

100 g x ( T_f - 20.72) ^oC= - (50.0 g x ( T_f - 100) ^oC

100 T_f - 2072 = - 50.0 T_f + 5000

150 T_f = 7072

T_f = 47.1 ^oC

Thus, the final temperature of the system is 47.1 ^oC