Respuesta :

skyluke89
The water density is 
[tex]d=1.0 g/mL = 1000 g/L[/tex]
And the mass of 1.40 L of water is
[tex]m=dV=(1000 g/L)(1.40 L)=1400 g[/tex]

The amount of heat needed to increase the temperature of the water by [tex]\Delta T[/tex] is given by
[tex]Q=m C_s \Delta T[/tex]
where m is the water mass, [tex]C_s = 4.18 J/g ^{\circ}C[/tex] is the water specific heat capacity and 
[tex]\Delta T=100.0 ^{\circ}C-20.0 ^{\circ}C = 80.0^{\circ}C[/tex] 
is the increase in temperature. If we substitute these numbers into the equation, we find
[tex]Q=(1400 g)(4.18 J/g^{\circ}C)(80.0^{\circ}C)=4.68 \cdot 10^5 J[/tex]

The heat required to warm 1.40 l of water from 20.0 °C to 100.0 °C is 468.608 KJ.

Given to us

  • The volume of the water, V = 1.40 l
  • Initial temperature, [tex]T_1[/tex] = 20 °C,
  • Final temperature, [tex]T_2[/tex] = 100 °C
  • Density of the water, ρ = 1.0 g/ml

Specific Heat of the Water

We know that the specific heat of the water is equal to  4184 J⋅kg−1⋅K−1.

Density of water

[tex]\rho = 1.0\ \dfrac{g}{ml} = 1.0\ \dfrac{\frac{1}{1000} kg}{\frac{1}{1000} l} = 1.0\ kg/l[/tex]

Mass of the water

[tex]Mass = Volume \times density\\\\m = V \times \rho\\m = 1.40 l\ \times 1.0\ kg/l\\m = 1.40\ kg[/tex]

Heat of the Water

We know  heat can be written as,

[tex]Q = m \times C \times \triangle T[/tex]

Substitute the values,

[tex]Q = 1.40 \times 4,184 \times (100 -20)\\Q = 1.40 \times 4,184 \times 80\\Q = 468,608\ J\\Q = 468.608\ KJ[/tex]

Hence, the heat required to warm 1.40 l of water from 20.0 ∘c to 100.0 ∘c is 468.608 KJ.

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