Question

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Part A but this time using an aluminum cylinder of equal mass? Assume that the initial temperatures of the metals (Tm) and the initial temperatures of the water (Ti) were the same. Specific heat of copper: 0.385 J/g*K Specific heat of aluminum: 0.900 J/g*K Group of answer choices The magnitudes of his q and ∆H for the copper trial would be higher than the aluminum trial. The magnitudes of his q and ∆H for both trials would be the same The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Answer 1

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

• initial temperature of metals, =  $T_m$
• initial temperature of water, = $T_i$
• specific heat capacity of copper, $C_p$ = 0.385 J/g.K
• specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
• both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

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