To find the amount of water in grams that was placed, we need to use the principle of energy conservation. In this situation, the heat lost by the hot water will be equal to the heat gained by the ice and the cold water.
Let's break down the steps to find the solution:
Step 1: Calculate the heat lost by the hot water.
- The initial temperature of the hot water is 55 degrees Celsius.
- The final temperature of the system (after reaching equilibrium) is 10 degrees Celsius.
- The specific heat capacity of water is 1 cal/(g.K).
Using the formula Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can calculate the heat lost by the hot water.
Q_hot = m_hot * c_water * ΔT_hot
ΔT_hot = 55 - 10 = 45 degrees Celsius
Step 2: Calculate the heat gained by the ice and the cold water.
- The final temperature of the system (after reaching equilibrium) is 10 degrees Celsius.
- The specific heat capacity of water is 1 cal/(g.K).
- The heat of fusion (melting) for ice is 80 cal/g.
For the ice, the heat gained is due to both the temperature change (from 0 to 10 degrees Celsius) and the phase change (from solid to liquid). We need to include both contributions.
Q_ice = Q_temp_change + Q_phase_change
Q_temp_change = m_ice * c_water * ΔT_ice
Q_phase_change = m_ice * heat_fusion_ice
ΔT_ice = 10 - 0 = 10 degrees Celsius
Step 3: Equate the heat lost and heat gained to find the mass of water.
Q_hot = Q_ice + Q_water
m_hot * c_water * ΔT_hot = (m_ice * c_water * ΔT_ice) + (m_water * c_water * ΔT_water)
Since specific heat capacity and temperature changes are the same for water, we can simplify the equation to:
m_hot * ΔT_hot = (m_ice * ΔT_ice) + (m_water * ΔT_water)
Now, we can substitute the known values and solve for m_water.
m_hot * ΔT_hot = (m_ice * ΔT_ice) + (m_water * ΔT_water)
m_water = (m_hot * ΔT_hot - m_ice * ΔT_ice) / ΔT_water
Substituting the values:
m_water = (100g * 45°C - 100g * 10°C) / (10°C - 55°C)
m_water = -3500g / -45°C
m_water = 77.78g
Therefore, the amount of water in grams that was placed is approximately 77.78 grams.