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Activity: A liquid of unknown specific heat at a temperature of 20^circ C was mixed with water at 80^circ C in a well-insulated container. The final temperature was measured to be 50^circ C and the combined mass of the two liquids was measured to be 240g. In a second experiment with both liquids at the same initial temperatures, 20 g less of the liquid of unknown specific heat was poured into the same amount of water as before. This time the equilibrium temperature was found to be 52^circ C. Determine the specific hear of the liquid. The specific heat of water is 4187J/Kg^circ C or 1kcal/kg^circ C.

Pergunta

Activity: A liquid of unknown specific heat at a temperature of 20^circ C
was mixed with water at 80^circ C
in a well-insulated container. The final temperature was measured to be
50^circ C and the
combined mass of the two liquids was measured to be 240g. In a second experiment with
both liquids at the same initial temperatures, 20 g less of the liquid of unknown specific heat
was poured into the same amount of water as before. This time the equilibrium temperature
was found to be 52^circ C.
Determine the specific hear of the liquid. The specific heat of water is
4187J/Kg^circ C or 1kcal/kg^circ C.

Activity: A liquid of unknown specific heat at a temperature of 20^circ C was mixed with water at 80^circ C in a well-insulated container. The final temperature was measured to be 50^circ C and the combined mass of the two liquids was measured to be 240g. In a second experiment with both liquids at the same initial temperatures, 20 g less of the liquid of unknown specific heat was poured into the same amount of water as before. This time the equilibrium temperature was found to be 52^circ C. Determine the specific hear of the liquid. The specific heat of water is 4187J/Kg^circ C or 1kcal/kg^circ C.

Solução

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KawanElite · Tutor por 8 anos

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To determine the specific heat of the unknown liquid, we can use the principle of conservation of energy. The heat lost by the hot liquid (water) will be equal to the heat gained by the cold liquid (unknown liquid).<br /><br />Let's denote the specific heat of the unknown liquid as \( c \) and its mass as \( m \). The specific heat of water is given as \( 4187 \, \text{J/kg}^\circ\text{C} \).<br /><br />In the first experiment:<br />- Initial temperature of water: \( 80^\circ\text{C} \)<br />- Final temperature: \( 50^\circ\text{C} \)<br />- Combined mass: \( 240 \, \text{g} \)<br /><br />In the second experiment:<br />- Initial temperature of water: \( 80^\circ\text{C} \)<br />- Final temperature: \( 52^\circ\text{C} \)<br />- Combined mass: \( 220 \, \text{g} \) (since 20g less of the unknown liquid was used)<br /><br />Using the formula for heat transfer:<br />\[ Q = mc\Delta T \]<br /><br />For the first experiment:<br />\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{initial, water}} - T_{\text{final}}) \]<br />\[ Q_{\text{unknown}} = m_{\text{unknown}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial, unknown}}) \]<br /><br />Since \( Q_{\text{water}} = Q_{\text{unknown}} \):<br />\[ m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{initial, water}} - T_{\text{final}}) = m_{\text{unknown}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial, unknown}}) \]<br /><br />For the second experiment:<br />\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{initial, water}} - T_{\text{final}}) \]<br />\[ Q_{\text{unknown}} = m_{\text{unknown}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial, unknown}}) \]<br /><br />Since \( Q_{\text{water}} = Q_{\text{unknown}} \):<br />\[ m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{initial, water}} - T_{\text{final}}) = m_{\text{unknown}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial, unknown}}) \]<br /><br />Let's solve these equations step by step.<br /><br />For the first experiment:<br />\[ m_{\text{water}} \cdot c_{\text{water}} \cdot (80 - 50) = m_{\text{unknown}} \cdot c \cdot (50 - 20) \]<br />\[ m_{\text{water}} \cdot 4187 \cdot 30 = m_{\text{unknown}} \cdot c \cdot 30 \]<br />\[ m_{\text{water}} \cdot 4187 = m_{\text{unknown}} \cdot c \]<br /><br />For the second experiment:<br />\[ m_{\text{water}} \cdot c_{\text{water}} \cdot (80 - 52) = m_{\text{unknown}} \cdot c \cdot (52 - 20) \]<br />\[ m_{\text{water}} \cdot 4187 \cdot 28 = m_{\text{unknown}} \cdot c \cdot 32 \]<br />\[ m_{\text{water}} \cdot 4187 \cdot 28 = m_{\text{unknown}} \cdot c \cdot 32 \]<br /><br />Now we have two equations:<br />1. \( m_{\text{water}} \cdot 4187 = m_{\text{unknown}} \cdot c \)<br />2. \( m_{\text{water}} \cdot 4187 \cdot 28 = m_{\text{unknown}} \cdot c \cdot 32 \)<br /><br />Divide the second equation by the first equation:<br />\[ \frac{m_{\text{water}} \cdot 4187 \cdot 28}{m_{\text{unknown}} \cdot c} = \frac{m_{\text{unknown}} \cdot c \cdot 32}{m_{\text{water}} \cdot 4187} \]<br />\[ 28 \cdot 32 = 32 \]<br />\[ 896 = 32 \]<br /><br />This simplifies to:<br />\[ c = \frac{
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