1. Dareo. [ m=2 mathrm(~m) t_(1)=14^circ mathrm(C) t_(2)=50^circ mathrm(C) Q_(1)=Q_(2) c-4200 (mathrm(Dm))/(k 2 cdot( )^circ mathrm(C)) (2=2,7 cdot 10^2 frac(2 m)/(22))(m_(2)-?) ]

To solve this problem, we need to use the formula for heat transfer:<br /><br />\[ Q = mc\Delta T \]<br /><br />where:<br />- \( Q \) is the heat transferred,<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 />- \( m = 2 \, \text{kg} \)<br />- \( t_1 = 14^\circ \text{C} \)<br />- \( t_2 = 50^\circ \text{C} \)<br />- \( Q_1 = Q_2 \)<br />- \( c = 4200 \, \text{J/(kg} \cdot \text{°C)} \)<br /><br />We need to find \( m_2 \).<br /><br />First, calculate the change in temperature (\( \Delta T \)):<br /><br />\[ \Delta T = t_2 - t_1 =^\circ \text{C} - 14^\text{C} = 36^\circ \text{C} \]<br /><br />Since \( Q_1 = Q_2 \), we can set up the equation:<br /><br />\[ Q_1 = Q_2 \]<br /><br />\[ m_1 c \Delta T_1 = m_2 c \Delta T_2 \]<br /><br />Given that \( m_1 = 2 \, \text{kg} \) and \( \Delta T_1 = 36^\circ \text{C} \), and \( \Delta T_2 = 36^\circ \text{C} \), we can simplify the equation:<br /><br />\[ 2 \, \text{kg} \cdot 4200 \, \text{J/(kg} \cdot \text{°C)} \cdot 36^\circ \text{C} = m_2 \cdot 4200 \, \text{J/(kg} \cdot \text{°C)} \cdot 36^\circ \text{C} \]<br /><br />\[ 2 \, \text{kg} = m_2 \]<br /><br />Therefore, the value of \( m_2 \) is \( 2 \, \text{kg} \).