The student used two different types of thermometer to measure the temperature

Physics

1 answer:

iris [78.8K]1 year ago

5 0

0.02 °C is the resolution of thermometer B while measuring temperature.

Thermometer A= 87.6

Thermometer B=87.4°C

resolution= 87.6- 87.4°C = 0.02 °C

Heat or cold are measured in terms of temperature. It can be defined in terms of a variety of arbitrary scales and shows how heat energy naturally moves from a hotter to a cooler substance (one at a lower temperature). Even if a match burns at a considerably greater temperature than an iceberg, an iceberg has a much higher overall heat energy. The energy of a thermodynamic system differs from temperature.

Temperature, along with pressure, density, and other related qualities, is referred to as an intense property—one that is independent of the amount of material being addressed—in contrast to broad characteristics like mass or volume.

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initial temperature of water, = $T_i$
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specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
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$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

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;

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