Assuming that the specic heat of salt is 0.22 Kcal/ (kg C),

could you please help me with this problem as soon as possible
01. Assuming that the specific heat of salt is 0.22 Kcal/ (kg °C), water is 1 Kcal/ (kg °C), and ice is 0.5 Kcal! (kg °C), calculate the specific heat of the water saltmixtures you made (use mass weighted averages). This means C (mixture) = m1C1 + m202/(m1+ m2) Here is one example: Mass of salt = 50 g Mass of water = 100 9 Plug the numbers into the equation to get C (mixture) = 0.74 Kcall (kg °C)