Question

Suppose 45.0 g of water at 85 °C is added to 105.0 g of ice at 0 °C. The molar heat of fusion of water is 6.01 kJ/mol, and the specific heat of water is 4.18 J/g °C. On the basis of these data, (a) what will be the final temperature of the mixture and (b) how many grams of ice will melt?

Answer 1

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