Respuesta :
Answer:
48.90°C is the final temperature of the mixture.
Explanation:
Mass of water with 25°C temperature = m
Volume of water = v = 270.0 mL
Density of the water = d = 1.00 g/mL
[tex]m=1.00 g/ml\times 270.0 mL=270.0 g[/tex]
Mass of water with 95°C temperature = M
Volume of water = V = 140.0 mL
Density of the water = d = 1.00 g/mL
[tex]M=1.00 g/ml\times 140.0 mL=140.0 g[/tex]
Specific heat of water = c
Heat absorbed by water with 25°C temperature = Q
Heat released by water with 25°C temperature = Q'
Final temperature after mixing = [tex]T_3[/tex]
Q = -Q' ( Law of conservation of heat)
[tex]m\times c\times (T_3-T_1)=-M\times c\times (T_3-T_2)[/tex]
[tex]270.0 g\times(T_3-25.00^oC)=-140.00 g\times (T_3-95.00^oC)[/tex]
[tex]T_3=48.90^oC[/tex]
48.90°C is the final temperature of the mixture.
The final temperature of the mixture of cold water and hot water is 48.9 ⁰C.
The given parameters;
- volume of the cold water = 270 ml
- temperature of the cold water = 25⁰C
- volume of the hot water, = 140 ml
- temperature of the hot water, = 95⁰C
- density of water, = 1 g/ml
The mass of the cold water is calculated as follows;
[tex]m = \rho \times V\\\\m_c = 1 \ g/ml \ \ \times \ \ 270 \ ml \\\\m_c = 270 \ g[/tex]
The mass of the hot water is calculated as follows;
[tex]m_h = 1 \ g/ml \ \ \times \ \ 140 \ ml\\\\m_h = 140 \ g[/tex]
The final temperature of the mixture is determined by applying the principle of conservation of energy;
[tex]m_c C \Delta \theta_c = m_h C \Delta \theta_h\\\\ m_c \Delta \theta_c = m_h \Delta \theta_h\\\\m_c (t - 25) = m_h(95 - t)\\\\270(t- 25) = 140(95-t)\\\\270t - 6750 = 13,300 - 140t\\\\410t = 20,050\\\\t = \frac{20,050}{410} \\\\t = 48.9 \ ^0C[/tex]
Thus, the final temperature of the mixture of cold water and hot water is 48.9 ⁰C.
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