100 grams of ice taken at a temperature of t ice = 0 degrees Celsius was mixed in a calorimeter with

100 grams of ice taken at a temperature of t ice = 0 degrees Celsius was mixed in a calorimeter with 100 grams of water, the temperature of which is t water = 70 degrees Celsius. Determine the final temperature of the mixture. The specific heat capacity of water is 4200 J / kg * K, the specific heat of melting of ice is 3.3 * 10–5 degrees J / kg, and the heat capacity of the calorimeter is neglected.

ml = 100 g = 0.1 kg.

tl = 0 ° C.

mw = 100 g = 0.1 kg.

tv = 70 ° C.

Cw = 4200 J / kg * ° C.

ql = 3.3 * 10 ^ 5 J / kg.

t -?

When ice is mixed with water, cooling water will give up the amount of heat Qw, which is determined by the formula: Qw = mw * Cw * (tv – t).

This amount of heat Qw will go to melting ice and heating water, which will be obtained by melting ice.

Qw = 0.1 kg * 4200 J / kg * ° C * (70 ° C – 0 ° C) = 29400 J.

The amount of thermal energy required to melt ice is determined by the formula: Ql = ql * ml.

Ql = 0.1 kg * 3.3 * 10 ^ 5 J / kg = 33000 J.

We see that the thermal energy that is released when the water is cooled is not enough to melt all the ice: Qw <Ql.

Answer: the temperature of the mixture will become t = 0 ° C.