26.40 g of ice at -10.3°C is placed in 37.37 g of water at95.1°C in a perfectly

Question

26.40 g of ice at -10.3°C is placed in 37.37 g of water at95.1°C in a perfectly insulated vessel. Assume that the molarheat capacities for H2O(s) and H2O(l) are 37.5 J/K/mol and 75.3J/K/mol, respectively, and the molar enthalpy of fusion for ice is6.01 kJ/mol. Calculate the final temperature. (You must answer inKelvin, not °C.) I'm not too sure where to start on this one. I tried doingqice=-qwater and I couldn't get it towork out because there is a Tfinal on both sides of theequation. 26.40 g of ice at -10.3°C is placed in 37.37 g of water at95.1°C in a perfectly insulated vessel. Assume that the molarheat capacities for H2O(s) and H2O(l) are 37.5 J/K/mol and 75.3J/K/mol, respectively, and the molar enthalpy of fusion for ice is6.01 kJ/mol. Calculate the final temperature. (You must answer inKelvin, not °C.) I'm not too sure where to start on this one. I tried doingqice=-qwater and I couldn't get it towork out because there is a Tfinal on both sides of theequation.

Explanation / Answer

We Know that :      According to Thermodynamicalrelationship :       Q = m x s x t    26.4 g / 18 g / mol x 37.5 J / mol-K x ( T- 262.7 ) K + 26.4 g / 18 g / mol x 6.01 x 103J = - 37.37 g / 18 g / mol x 75.3 J / mol-K x (T-368.1)K       T = 298.960 K

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