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: 4.18J/g^circ C and ice: 2.108J/g^circ C) a. 20.064 J b. 14150J c. 200641 d. 10118.4 J

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 />\[ Q = 160 \, \text{g} \times 2.108 \, \text{J/g°C} \times 30^{\circ}C \]<br /><br />\[ Q = 160 \times 2.108 \times 30 \]<br /><br />\[ Q = 10064.8 \, \text{J} \]<br /><br />Therefore, the heat gain by the ice is 10064.8 J. The closest answer is:<br /><br />d. 10118.4 J