Cooling Down Hot Water with Ice
First, we need to calculate the heat lost by the water, which is given by Q_water = m_water * C_water * (800 - 50), where m_water is the mass of water and C_water is its specific heat capacity. Since we know that the mass of the water is 1,2 kg, and the specific heat capacity of water is 4.186 J/(g°C), we can plug these values into the equation to get Q_water = 1,2 * 4.186 * (800 - 50) = 3 750,96 J.
Next, we need to calculate the heat gained by the ice, which is given by Q_ice = m_ice * C_ice * (00 - 50), where m_ice is the mass of ice and C_ice is its specific heat capacity. However, since the ice is at 00С, it also needs to go through a phase change to become water at 50С. This requires an additional amount of heat, which is given by Q_phase = m_ice * L_f, where L_f is the latent heat of fusion for ice, which is 334 J/g. Thus, the total heat gained by the ice is Q_ice = m_ice * (C_ice * (00 - 00050) + L_f).
We can now combine the two equations together and solve for m_ice: 3 750,96 J = m_ice * (4.186 * (800 - 50) + 334)
m_ice = 3 75096 / (4.186 * (800 - 50) + 334) = 0,82 kg.
Therefore, you will need 0,82 kg of ice (or about 820 grams) to cool down the water from 800C to 50C.
Just be careful not to accidentally drop your phone in the water while you're adding the ice. Accidents like that can really put a damper on your day. Good luck with your experiment!
