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Answer :
120.6 g. of ice at 0°C must be added to lower the temperature of the tea to 16°C . Option C is correct.
Mass of water (m_water) = 1 liter = 1000 g (since density of water is 1 g/mL)
Initial temperature of water (T_water_initial) = 35°C
Final temperature of the mixture (T_final) = 16°C
Specific heat of water (c_water) = 1 cal/g.°C
Heat lost by water (Q_water) can be calculated using the formula: Q_water = m_water * c_water * (T_water_initial - T_final)
Heat required to melt the ice (Q_melt)
Heat required to raise the temperature of the melted ice to the final temperature (Q_ice)
Latent heat of fusion of ice (L_f) = 79.7 cal/g (heat required to melt 1 gram of ice)
Mass of ice (m_ice) is what we're solving for.
Q_melt = m_ice * L_f
Specific heat of water (c_water) applies again for the temperature change of melted ice (T_ice = 0°C, which is the temperature of ice before melting).
Q_ice = m_ice * c_water * (T_final - T_ice)
In a closed system (no heat transfer to the surroundings), the heat lost by the water will be equal to the heat gained by the melted ice to reach the final temperature.
Therefore, Q_water = Q_melt + Q_ice
:m_water * c_water * (T_water_initial - T_final) = m_ice * L_f + m_ice * c_water * (T_final - T_ice)
m_ice (L_f + c_water * T_final) = m_water * c_water * (T_water_initial - T_final)
m_ice = (m_water * c_water * (T_water_initial - T_final)) / (L_f + c_water * T_final)
m_water = 1000 g
c_water = 1 cal/g.°C
T_water_initial = 35°C
T_final = 16°C
L_f = 79.7 cal/g m_ice = (1000 g * 1 cal/g.°C * (35°C - 16°C)) / (79.7 cal/g + 1 cal/g.°C * 16°C) m_ice
≈ 120.6 g
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