160 grams ice at 0^circ C with the final temperature of 30^circ C . What is the heat gain by ice? (Specific Heat Capacity; Water: 4 . 18 J/g^circ C and ice: 2.108J/g^circ C) a. 14150J b. 20.064 J c. 20064J d. 101184J

To calculate the heat gain by the ice, we can use the formula:<br /><br />\[ Q = m \cdot c \cdot \Delta T \]<br /><br />Where:<br />- \( Q \) is the heat gain<br />- \( m \) is the mass of the ice<br />- \( c \) is the specific heat capacity of ice<br />- \( \Delta T \) is the change in temperature<br /><br />Given:<br />- \( m = 160 \) grams<br />- \( c = 2.108 \) J/g°C<br />- \( \Delta T = 30^{\circ}C - 0^{\circ}C = 30^{\circ}C \)<br /><br />Now, plug in the values:<br /><br />\[ Q = 160 \cdot 2.108 \cdot 30 \]<br /><br />\[ Q = 10064 \cdot 30 \]<br /><br />\[ Q = 101184 \text{ J} \]<br /><br />Therefore, the correct answer is:<br /><br />d. 101184 J