149g water at 50^circ C is added to 104 gice at 0^circ C. Final temperature of the mixture is 26^circ C. Calculate the heat lost by water?(Specific Heat Capacity; Water: 4.18J/g^circ C and ice: 2.108J/g^circ C) a. -14948J b. 14.948 KJ -18.18KJ d. 18183J

To calculate the heat lost by water, we can use the formula:<br /><br />\[ Q = m \cdot c \cdot \Delta T \]<br /><br />where:<br />- \( Q \) is the heat lost or gained,<br />- \( m \) is the mass of the substance,<br />- \( c \) is the specific heat capacity of the substance,<br />- \( \Delta T \) is the change in temperature.<br /><br />Given:<br />- Mass of water (\( m_{\text{water}} \)) = 149 g<br />- Specific heat capacity of water (\( c_{\text{water}} \)) = 4.18 J/g°C<br />- Initial temperature of water (\( T_{\text{initial, water}} \)) = 50°C<br />- Final temperature of the mixture (\( T_{\text{final}} \)) = 26°C<br /><br />The change in temperature for water (\( \Delta T_{\text{water}} \)) is:<br /><br />\[ \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} = 26°C - 50°C = -24°C \]<br /><br />Now, we can calculate the heat lost by water:<br /><br />\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]<br />\[ Q_{\text{water}} = 149 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (-24°C) \]<br />\[ Q_{\text{water}} = 149 \cdot 4.18 \cdot (-24) \]<br />\[ Q_{\text{water}} = -14948 \, \text{J} \]<br /><br />Therefore, the heat lost by water is \( -14948 \, \text{J} \).<br /><br />The correct answer is:<br />a. \( -14948 \, \text{J} \)