A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0 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 $0.18 \mathrm { cal } / \mathrm { g } \cdot { } ^ { \circ } \mathrm { C }$ and that of water is $1.0 \mathrm { cal } / \mathrm { g } \cdot \mathrm { C } ^ { \circ }$

Question

A glass beaker of unknown mass contains  74.0 ml of water. The system absorbs  2000.0 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 $0.18 \mathrm { cal } / \mathrm { g } \cdot { } ^ { \circ } \mathrm { C }$ and that of water is $1.0 \mathrm { cal } / \mathrm { g } \cdot \mathrm { C } ^ { \circ }$

Quizplus · Accepted Answer

The total heat absorbed is distributed between heating the water and the beaker. First, calculate the heat absorbed by the water: $Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T$, where $m_{water} = 74.0 \, g$ (since 1 ml of water = 1 g), $c_{water} = 1.0 \, cal/g \cdot ^\circ C$, and $\Delta T = 20.0 \, ^\circ C$. So, $Q_{water} = 74.0 \cdot 1.0 \cdot 20.0 = 1480.0 \, cal$. The remaining heat goes to the beaker: $Q_{beaker} = 2000.0 - 1480.0 = 520.0 \, cal$. Using $Q_{beaker} = m_{beaker} \cdot c_{beaker} \cdot \Delta T$, where $c_{beaker} = 0.18 \, cal/g \cdot ^\circ C$, solve for $m_{beaker}$: $520.0 = m_{beaker} \cdot 0.18 \cdot 20.0$, which gives $m_{beaker} = 520.0 / (0.18 \cdot 20.0) = 520.0 / 3.6 = 144.44 \, g$, which is closest to 140 g.

A Glass Beaker of Unknown Mass Contains 74 20.0∘C20.0 ^ { \circ } \mathrm { C }20.0∘C

A Glass Beaker of Unknown Mass Contains 74 $20.0 ^ { \circ } \mathrm { C }$