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?

Question

Karson Delgado

Chemistry

16 December, 05:25

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?

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Answers (1)

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Elise Evans · Answer 1

A. Final Temp = 36.428C and

47.9g ice will melt

Explanation:

Given the following dа ta:

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