Answer:
A. Final Temp = 36.428C and
47.9g ice will melt
Explanation:
Given the following data:
Mass of water (M1) = 45.0g = 0.045kg
Temperature (T1) = 85C = 358k
Mass of ice (M2) = 105g = 0.105kg
Temperature (T2) = 0c = 273k
Specific heat of water (C) = 4.18j/gC = 0.00418kj/kgc
Molar heat of fusion of water = 6.01kj/mol
Therefore, heat required (q) = MCT
M1C(T1-T2) = M2C∆T
By putting the data we have
0.045×0.00418×(358-273) = 0.105×0.00418×∆T
∆t = 0.045×0.00418×85/0.105×0.00418
∆t = 36.428C
Gram of ice that would melt would be 47.9g
