Question

a 95.0 g sample of copper (c = 0.20 j/°c·g) is heated to 82.4°c and then placed in a container of water (c = 4.18 j/°c·g) at 22.0°c. the final temperature of the water and the copper is 25.1°c. what was the mass of the water in the original container? assume that all heat lost by the copper is gained by the water. use the formulas below to help in your problem - solving.
-q_metal = q_water
-c_m m_m δt_m = c_w m_w δt_w
0.246 g h₂o
4.73 g h₂o
84.0 g h₂o
36.700 g h₂o

Explanation:

Step1: Calculate the temperature change of copper

The initial temperature of copper \(T_{i,cu}=82.4^{\circ}C\) and the final temperature \(T_f = 25.1^{\circ}C\).
\(\Delta T_{cu}=T_f - T_{i,cu}=25.1 - 82.4=- 57.3^{\circ}C\)

Step2: Calculate the temperature change of water

The initial temperature of water \(T_{i,w}=22.0^{\circ}C\) and the final temperature \(T_f = 25.1^{\circ}C\).
\(\Delta T_{w}=T_f - T_{i,w}=25.1 - 22.0 = 3.1^{\circ}C\)

Step3: Rearrange the heat - transfer formula for the mass of water

We know that \(-c_{cu}m_{cu}\Delta T_{cu}=c_{w}m_{w}\Delta T_{w}\).
We want to solve for \(m_{w}\), so \(m_{w}=\frac{-c_{cu}m_{cu}\Delta T_{cu}}{c_{w}\Delta T_{w}}\)
Given \(c_{cu}=0.20\ J/^{\circ}C\cdot g\), \(m_{cu} = 95.0\ g\), \(c_{w}=4.18\ J/^{\circ}C\cdot g\)
Substitute the values:

$$
LATEXBLOCK0
$$

Answer:

\(84.0\ g\ H_2O\) (the third option)