Question

The temperature of 1.94 kg of water is 34 °C. To cool the water, ice at 0°C is added to it. The desired final temperature of the water is 11 °C. The latent heat of fusion for water is 33.5 × 10^4 J/kg, and the specific heat capacity of water is 4186 J/(kg·C°). Ignoring the container and any heat lost or gained to or from the surroundings, determine how much mass m of ice should be added?

Answer 1

Answer:

0.4042 kg

Explanation:

Please see attachment

Answer 2

Answer:

0.058 kg or 58 g

Explanation:

The amount of heat energy required to cool 1.94 kg of water from 34 Celcius to 11 Celcius degrees is:

$H = mc\Delta t$

where m = 1.94 kg is the water mass, c = 4186 J/kg.C is water specific heat and $\Delta t = t_2 - t_1 = 34 - 11 = 23^o C$

Hence H = 1.94 * 4186 * 23 = 186779.32  J

The required mass of ice to absorb this amount of heat would be

$m_i = \frac{H}{L_f} = \frac{186779.32}{33.5*10^4} = 0.058 kg = 58 g$