A glass beaker of unknown mass contains $74.0 \mathrm { ml }$ of water. The system absorbs $2000.0 \mathrm { cal }$ of heat 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°.
A)140 g
B)560 g
C)540 g
D)270,000 g

Question

Quizplus · Accepted Answer

The heat absorbed by the water is:$$Q_w = m_w c_w \Delta T = (74.0 \mathrm { g })(1.0 \mathrm { cal/g } \cdot ^\circ C)(20.0 ^\circ C) = 1480 \mathrm { cal }$$The heat absorbed by the glass is:$$Q_g = m_g c_g \Delta T = m_g (0.18 \mathrm { cal/g } \cdot ^\circ C)(20.0 ^\circ C) = 36.0 m_g \mathrm { cal }$$The total heat absorbed by the system is:$$Q = Q_w + Q_g = 1480 \mathrm { cal } + 36.0 m_g \mathrm { cal } = 2000.0 \mathrm { cal }$$Solving for $m_g$, we get:$$m_g = \frac{2000.0 \mathrm { cal } - 1480 \mathrm { cal }}{36.0 \mathrm { cal/g } \cdot ^\circ C} = 140 \mathrm { g }$$

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

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