Chemistry I20) An ice cube at 0C was dropped into 30.0 g of water in a cup at 45.0C. At the instant that all of the ice was melted, the temperature of the water in the cup was 19.5C. What was the mass of ice cube? 9.6 g 6.9 g 9.8 g 8.9 gPrevious Question Next QuestionDo not use the 'Forward' or 'Back' button on your browser.
Question
Chemistry I20) An ice cube at 0C was dropped into 30.0 g of water in a cup at 45.0C. At the instant that all of the ice was melted, the temperature of the water in the cup was 19.5C. What was the mass of ice cube? 9.6 g 6.9 g 9.8 g 8.9 gPrevious Question Next QuestionDo not use the 'Forward' or 'Back' button on your browser.
Solution
To solve this problem, we need to use the concept of heat transfer. The heat gained by the ice to melt and raise its temperature to the final temperature is equal to the heat lost by the water to cool down to the final temperature.
First, let's calculate the heat lost by the water:
Q_water = m_water * c_water * ΔT_water where: m_water = mass of the water = 30.0 g c_water = specific heat capacity of water = 4.18 J/(g°C) ΔT_water = change in temperature of the water = initial temperature - final temperature = 45.0°C - 19.5°C = 25.5°C
Q_water = 30.0 g * 4.18 J/(g°C) * 25.5°C = 3196.95 J
Next, let's calculate the mass of the ice. The heat gained by the ice is used to first melt the ice and then raise its temperature to the final temperature. This can be expressed as:
Q_ice = m_ice * Hf_ice + m_ice * c_water * ΔT_ice where: m_ice = mass of the ice (which we're trying to find) Hf_ice = heat of fusion of ice = 334 J/g ΔT_ice = change in temperature of the ice = final temperature - initial temperature = 19.5°C - 0°C = 19.5°C
Since the heat lost by the water is equal to the heat gained by the ice, we can set Q_water equal to Q_ice and solve for m_ice:
3196.95 J = m_ice * 334 J/g + m_ice * 4.18 J/(g°C) * 19.5°C
Solving this equation for m_ice gives the mass of the ice cube.
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