Question

1. a piece of gold with a mass of 21.0 g was heated to 100.0 °c and placed inside a coffee cup calorimeter. the temperature of the water inside the calorimeter increased from 23.8 °c to 25.5 °c. the specific heat capacity of water is 4.184 j g⁻¹ °c⁻¹ and the specific heat capacity of gold is 0.129 j g⁻¹ °c⁻¹. calculate the mass of water (in grams, g) contained inside the calorimeter. assume that no heat is lost to the coffee cup calorimeter or the surroundings.

a) 28.4 g
b) 21.0 g
c) 16.0 g
d) 39.5 g
e) 32.9 g

Explanation:

Step1: Set up heat - transfer equation

According to the law of conservation of energy, the heat lost by gold is equal to the heat gained by water, i.e., $q_{gold}=-q_{water}$. The heat - transfer formula is $q = mc\Delta T$. So, $m_{gold}c_{gold}\Delta T_{gold}=-m_{water}c_{water}\Delta T_{water}$.

Step2: Calculate $\Delta T$ values

For gold, $\Delta T_{gold}=T_{final}-T_{initial}=25.5^{\circ}C - 100.0^{\circ}C=- 74.5^{\circ}C$. For water, $\Delta T_{water}=25.5^{\circ}C - 23.8^{\circ}C = 1.7^{\circ}C$.

Step3: Rearrange the equation to solve for $m_{water}$

We know that $m_{gold} = 21.0\ g$, $c_{gold}=0.129\ J\ g^{-1}\ ^{\circ}C^{-1}$, $c_{water}=4.184\ J\ g^{-1}\ ^{\circ}C^{-1}$. Rearranging $m_{gold}c_{gold}\Delta T_{gold}=-m_{water}c_{water}\Delta T_{water}$ gives $m_{water}=\frac{m_{gold}c_{gold}\Delta T_{gold}}{-c_{water}\Delta T_{water}}$.

Step4: Substitute the values

$m_{water}=\frac{21.0\ g\times0.129\ J\ g^{-1}\ ^{\circ}C^{-1}\times(-74.5^{\circ}C)}{-4.184\ J\ g^{-1}\ ^{\circ}C^{-1}\times1.7^{\circ}C}$
$m_{water}=\frac{21.0\times0.129\times74.5}{4.184\times1.7}$
$m_{water}=\frac{21.0\times9.6105}{7.1128}$
$m_{water}=\frac{201.8205}{7.1128}\approx28.4\ g$

Answer:

A. 28.4 g