Answer:
338 K
Explanation:
[tex]m_{1}[/tex] = 2.5 kg, c = 4189 J/(kg K), [tex]T_{1}[/tex] = 13.5 [tex]^{o} C[/tex], [tex]T_{2}[/tex] = 22.5 [tex]^{o} C[/tex]
[tex]m_{2}[/tex] = 0.5 kg, [tex]T_{3}[/tex] = 20 [tex]^{o} C[/tex]
Heat loss by hotter water = heat gained by cooler water
[tex]m_{1}[/tex]cΔT = [tex]m_{2}[/tex]cΔT
[tex]m_{1}[/tex]c([tex]T_{2}[/tex] - [tex]T_{1}[/tex]) = [tex]m_{2}[/tex]c([tex]T_{4}[/tex] - [tex]T_{3}[/tex])
2.5 x 4189 x (22.5 - 13.5) = 0.5 x 4189 x ([tex]T_{4}[/tex] - 20)
2.5 x 4189 x 9 = 2094.5 ([tex]T_{4}[/tex] - 20)
94252.5 = 2094.5[tex]T_{4}[/tex] - 41890
94252.5 + 41890 = 2094.5[tex]T_{4}[/tex]
136142.5 = 2094.5[tex]T_{4}[/tex]
[tex]T_{4}[/tex] = [tex]\frac{136142.5}{2094.5}[/tex]
= 65
[tex]T_{4}[/tex] = 65 [tex]^{o} C[/tex]
But,
θ K = 273 + θ[tex]^{o} C[/tex]
= 273 + 65
= 338
The final temperature of the water is 338 K.
The value of the temperature in kelvin is 338 K.
The given parameters;
The final temperature of the water is calculated as follows as shown below;
[tex]m_1C\Delta \theta_1 = m_2C\Delta \theta_2\\\\m_1\Delta \theta_1 = m_2\Delta \theta_2\\\\2.5(22.5 - 13.5) = 0.5(t_4- 20)\\\\22.5 = 0.5t_4 - 10\\\\0.5t_4 = 32.5\\\\t_4 = \frac{32.5}{0.5} \\\\t_4 = 65 \ ^0C[/tex]
The value of the temperature in kelvin is calculated as;
[tex]t_4 = 273 + 65\\\\t_4 = 338 \ K[/tex]
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