Answer:
m-Ice = 23.85g
Explanation:
The heat absorbed for the cold water (Heat of the increase of water + Heat of ice melting) is equal to the heat released for the water at 75°C. The equation is:
m-Ice-*C*ΔT + Hf*-m-Ice = m-water*C*ΔT
<em>Where m-Ice- is our incognite</em>
<em>C is specific heat of water = 4.18J/g°C</em>
<em>ΔT is change in temperature = 25.0°C - 0.0°C = 25.0°C</em>
<em>Hf is enthalpy of fusion of water = 333.6J/g</em>
<em>m-water = 50.0g</em>
<em>ΔT is change in temperature = 75.0°C - 25.0°C = 50.0°C</em>
<em />
m-Ice-*4.18J/g°C*25°C + 333.6J/g*-m-Ice = 50.0g*4.18J/g°C*50.0°C
104.5 m-Ice- + 333.6 m-Ice- = 10450
438.1 m-Ice = 10450
<h3>m-Ice = 23.85g</h3>
