Answer to Question #99596 in Classical Mechanics for Ibrahim Mohamed

Question

How many 20g ice cubes, whose initial temperature is -10 C must be added to 1.0L of hot tea whose initial temperature is 90 C, in order that the final mixture be 10 C. Assume that all ice melt at the final mixture and specific heat of tea is same as that of water.
Lf of ice = 333 Kj/kg
Specific heat of water = 4190
Specific heat of ice = 2100
Assume that tea has same properties as water and 1L of water has 1Kg mass

Answer 1

Let us use notation:

$m_w = 1 kg$ - mass of the water, $m_i = 0.02 kg$ - mass of one ice cube,

$T_i = -10^{\circ} C$ - initial temperature of the ice cube, $T_t = 90^{\circ}C$ - initial temperature of the tea, $T = 10^{\circ}C$- final temperature,

$c_w = 4190 \frac{J}{kg \cdot ^{\circ}C}$ - specific heat of water, $c_i = 2100 \frac{J}{kg \cdot ^{\circ}C}$ - specific heat of ice, $\lambda_i = 333 \frac{K J}{kg}$ - specific latent heat of ice.

Lets assume, we need $n$ ice cubes. The energy, gained from cooling the tea from $T_t$ to $T$ is used to heat the ice cubes from $T_i$ to melting point of ice $0^{\circ}C$, melt them, and warm the obtained liquid to $T$:

$c_w m_w (T_t - T) = n c_i m_i (0 - T_i) + \lambda n m_i + n m_i c_w (T - 0)$, from where $n = \frac{c_w m_w (T_t - T)}{m_i(c_w T + \lambda - c_i T_i)}$ . Plugging in given values and calculating, obtain $n \approx 42$. Hence, we need 42 ice cubes.