Respuesta :
[text]Solution:\\ Heat blance is gained by 50g =Heat balance lost by 25g of water\\ So,\\ \(M1 \times temp + M2 \times temp/M1+M2)\\\\\\\frac{(50.0g\times25.0^{.}c)+(23.0g\times57.0^{.}c)}{50.0g+23.0g} =35.1^{.}c}[/tex]
Answer: The final temperature of the water is 6.28°C
Explanation:
When liquid water is mixed with water, the amount of heat released by liquid water will be equal to the amount of heat absorbed by water.
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex] ......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of liquid water = 50.0 g
[tex]m_2[/tex] = mass of water = 35.0 g
[tex]T_{final}[/tex] = final temperature = ?°C
[tex]T_1[/tex] = initial temperature of liquid water = 25.0°C
[tex]T_2[/tex] = initial temperature of water = 51.0°C
[tex]c[/tex] = specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
[tex]50.0\times 4.186\times (T_{final}-25)=-[35.0\times 4.186\times (T_{final}-51)][/tex]
[tex]T_{final}=6.28^oC[/tex]
Hence, the final temperature of the water is 6.28°C
