Answer to Question #148838 in Molecular Physics | Thermodynamics for Kyra Nameth

Question

If 15.0 g of Ice at -25.0°C is mixed with a mass of antifreeze initially at -3.00°C. If the final temperature of the mixture is -10.0°C, how much antifreeze is present (Cantifreeze=2.24 x 102 J/KgoC)?

Answer 1

$m_1 = 15.0g = \frac{15}{1000}kg\\ m_2 = ?\\ c_1 = 4.2×10³J/kg.K\\ c_2 = 2.24×10²J/kg.K\\ T_1 = -25.0°C\\ T_2 = -3.0°C\\ T_3 = -10.0°C$

from,

Heat lost = Heat gained

$m_1c_1∆T = m_2c_2∆T\\ m_1c_1(T3-T1) = m_2c_2(T2-T3)$

$\frac{15}{1000}×4.2×10³×(-10+25) = m_2×2.24×10²×(-3+10)\\ 63×1000×\frac{15}{1000} = m2×15.68×100$

$m_2 = \dfrac{63×1000×15}{15.68×100×1000}$

$m2 = 0.603kg = 603g$