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Five grams of 20°C water is poured onto a 6-g block of ice that has an initial temperature of -15°C.
What is the final temperature if the effect of the surroundings can be neglected? .( specific heat of
ice = 0.50 cal/g.0C)

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To find the final temperature of the system after the water is poured onto the ice, we can use the principle of conservation of energy. The heat lost by the water as it cools down will be equal to the heat gained by the ice as it melts and increases in temperature.
First, we calculate the heat lost by the water:
Q_water = m_water * c_water * ΔT
Q_water = 5g * 1 cal/g°C * (T_final - 20°C)
Next, we calculate the heat gained by the ice:
Q_ice = m_ice * L_f + m_ice * c_ice * ΔT
Q_ice = 6g * 80 cal/g + 6g * 0.50 cal/g°C * (T_final - (-15°C))
Because the heat lost by the water is equal to the heat gained by the ice:
5g * 1 cal/g°C * (T_final - 20°C) = 6g * 80 cal/g + 6g * 0.50 cal/g°C * (T_final - (-15°C))
Solving for T_final:
5(T_final - 20) = 480 + 3(T_final + 15)
5T_final - 100 = 480 + 3T_final + 45
5T_final - 3T_final = 480 + 45 + 100
2T_final = 625
T_final = 625 / 2
T_final = 312.5°C
Therefore, the final temperature of the system would be approximately 312.5°C.

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