If you have 0.361 m 3 0.361 m3 of water at 25.0 ∘ C 25.0 ∘C and add 0.127 m 3 0.127 m3 of water at 95.0 ∘ C 95.0 ∘C , what is the final temperature of the mixture? Use 1000 kg/m 3 1000 kg/m3 as the density of water at any temperature.

Respuesta :

Answer:

Therefore the final temperature of the mixture will be 43.22°C

Explanation:

Density is given as a function of mass in a given volume, mathematically it can be expressed as

[tex]\rho = \frac{m}{V}[/tex]

Where,

m = mass

V = Volume

we have,

[tex]m = \rho V[/tex]

For state 1

we have that the mass is

[tex]m_1 = (1000kg/m^3)(0.361m^3) = 361kg[/tex]

For state 2

[tex]m_2 = (1000kg/m^3)(0.127m^3) = 127kg[/tex]

from calorimetry we know that heat change is given under

[tex]Q = mc_p\Delta T[/tex]

For energy conservation then,

[tex]m_1c_p\Delta T = m_2c_p\Delta T[/tex]

Since the specific heat is the same for the fluid then,

[tex]m_1\Delta T = m_2\Delta T[/tex]

[tex](361) (T-25\°C) = (127) (95\°C-T)T = 43.22\°C[/tex]

Therefore the final temperature of the mixture will be 43.22°C