NaHo: m_(1)=0,2Kr m_(2)=0,8Kr c_(1)=c_(2)=4200(甲times K)/(Krcdot ^circ )C t_(1)=100^circ C t_(2)=25^circ C t=40^circ C Q_(2)- ? Q_(1)-?

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, and<br />- $\Delta t$ is the change in temperature.<br /><br />Given:<br />- $m_{1} = 0.2 \, \text{kg}$<br />- $m_{2} = 0.8 \, \text{kg}$<br />- $c_{1} = c_{2} = 4200 \, \text{J/(kg} \cdot \text{°C)}$<br />- $t_{1} = 100 \, \text{°C}$<br />- $t_{2} = 25 \, \text{°C}$<br />- $t = 40 \, \text{°C}$<br /><br />We need to find $Q_{1}$ and $Q_{2}$.<br /><br />First, let's calculate the change in temperature for each substance:<br />- $\Delta t_{1} = t - t_{1} = 40 \, \text{°C} - 100 \, \text{°C} = -60 \, \text{°C}$<br />- $\Delta t_{2} = t - t_{2} = 40 \, \text{°C} - 25 \, \text{°C} = 15 \, \text{°C}$<br /><br />Now, we can calculate $Q_{1}$ and $Q_{2}$:<br />- $Q_{1} = m_{1} \cdot c_{1} \cdot \Delta t_{1} = 0.2 \, \text{kg} \cdot 4200 \, \text{J/(kg} \cdot \text{°C)} \cdot (-60 \, \text{°C}) = -50400 \, \text{J}$<br />- $Q_{2} = m_{2} \cdot c_{2} \cdot \Delta t_{2} = 0.8 \, \text{kg} \cdot 4200 \, \text{J/(kg} \cdot \text{°C)} \cdot 15 \, \text{°C} = 50400 \, \text{J}$<br /><br />Therefore, $Q_{1 -50400 \, \text{J}$ and $Q_{2} = 50400 \, \text{J}$.