A glass beaker of unknown mass contains $74.0 \mathrm { ml }$ of water.The system absorbs $2000.0 \mathrm { cal }$ of thermal energy and the temperature rises $20.0 ^ { \circ } \mathrm { C }$ as a result.What is the mass of the beaker? The specific heat of glass is 0.18 cal/g ∙ °C,and that of water is 1.0 cal/g ∙ C°.

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Quizplus · Accepted Answer

The total thermal energy absorbed by the system is given as $2000.0 \mathrm { cal }$. This energy is used to raise the temperature of both the water and the glass beaker. The amount of energy required to raise the temperature of the water can be calculated using the formula $q = mc\Delta T$, where $m$ is the mass of the water, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For water, $m = 74.0 \mathrm { g }$ (since $1 \mathrm { ml } = 1 \mathrm { g }$), $c = 1.0 \mathrm { cal/g \cdot ^\circ C }$, and $\Delta T = 20.0 ^\circ \mathrm { C }$. Thus, the energy absorbed by the water is $q_{\text{water}} = 74.0 \mathrm { g } \times 1.0 \mathrm { cal/g \cdot ^\circ C } \times 20.0 ^\circ \mathrm { C } = 1480.0 \mathrm { cal }$.The remaining energy is absorbed by the glass beaker, which is $2000.0 \mathrm { cal } - 1480.0 \mathrm { cal } = 520.0 \mathrm { cal }$. Using the same formula for the glass beaker, where $c = 0.18 \mathrm { cal/g \cdot ^\circ C }$ and $\Delta T = 20.0 ^\circ \mathrm { C }$, we can solve for the mass of the beaker, $m_{\text{beaker}}$, as follows: $520.0 \mathrm { cal } = m_{\text{beaker}} \times 0.18 \mathrm { cal/g \cdot ^\circ C } \times 20.0 ^\circ \mathrm { C }$. Solving for $m_{\text{beaker}}$ gives $m_{\text{beaker}} = \frac{520.0 \mathrm { cal }}{0.18 \mathrm { cal/g \cdot ^\circ C } \times 20.0 ^\circ \mathrm { C }} = 144.44 \mathrm { g }$, which rounds to $140 \mathrm { g }$ when considering the options provided.

A Glass Beaker of Unknown Mass Contains 74.0ml74.0 \mathrm { ml }74.0ml

A Glass Beaker of Unknown Mass Contains $74.0 \mathrm { ml }$