How much heat must be added to a 8.0-kg block of ice at -8°C to change it to water at $14 ^ { \circ } \mathrm { C } \text { ? }$ The specific heat of ice is 2050 J/kg ∙ C°, the specific heat of water is 4186 J/kg ∙ C°, the latent heat of fusion of ice is 334,000 J/kg, and 1 cal = 4.186 J.
A) 780 kcal
B) 140 kcal
C) 180 kcal
D) 810 kcal
E) 730 kcal

Question

Quizplus · Accepted Answer

To calculate the total heat required, we need to consider three steps: heating the ice from -8°C to 0°C, melting the ice, and then heating the water from 0°C to 14°C. 1. Heating the ice:    $ Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T = 8 \, \text{kg} \cdot 2050 \, \text{J/kg°C} \cdot 8 \, \text{°C} = 131,200 \, \text{J} $.2. Melting the ice:   $ Q_2 = m \cdot L_f = 8 \, \text{kg} \cdot 334,000 \, \text{J/kg} = 2,672,000 \, \text{J} $.3. Heating the water:   $ Q_3 = m \cdot c_{\text{water}} \cdot \Delta T = 8 \, \text{kg} \cdot 4186 \, \text{J/kg°C} \cdot 14 \, \text{°C} = 469,888 \, \text{J} $.Total heat $ Q = Q_1 + Q_2 + Q_3 = 131,200 + 2,672,000 + 469,888 = 3,273,088 \, \text{J} $.Convert to kcal: $ 3,273,088 \, \text{J} \times \frac{1 \, \text{kcal}}{4186 \, \text{J}} \approx 781.8 \, \text{kcal} $.Rounded to the nearest whole number, this is approximately 780 kcal.

How Much Heat Must Be Added to a 8 14∘C ? 14 ^ { \circ } \mathrm { C } \text { ? }14∘C ?

How Much Heat Must Be Added to a 8 $14 ^ { \circ } \mathrm { C } \text { ? }$