Question

A chemist heats 37.35 g of lead to 67.71 °C , then places the metal sample in the cup of water shown in the interactive. Calculate the final temperature of the water. The specific heat of lead is 0.130 J/g⋅°C and the specific heat of water is 4.184 J/g⋅°C .

mass of water: 60.000g
initial temperature of water: 25.00 C

Answer 1

Answer:

$T_{eq}=25.81\°C$

Explanation:

Hello!

In this case, since the calorimetry problems involving how the heat released by a hot substance is gained by the cold substance when they are placed in contact until they reach the thermal equilibrium is mathematically defined via:

$Q_{hot}=-Q_{cold}$

In this problem, it is seen that the lead is hot and the water cold, so we can write:

$Q_{lead}=-Q_{water}\\\\m_{lead}C_{lead}(T_{eq}-T_{lead})=-m_{water}C_{water}(T_{eq}-T_{water})$

Thus, since it is asked for the final temperature, or equilibrium temperature, we solve for it as shown below:

$T_{eq}=\frac{m_{lead}C_{lead}T_{lead}+m_{water}C_{water}T_{water}}{m_{lead}C_{lead}+m_{water}C_{water}}$

Then, by plugging in the known data we obtain:

$T_{eq}=\frac{37.35g*0.130\frac{J}{g\°C} *67.71\°C+60.000g*4.184\frac{J}{g\°C}*25.00\°C}{37.35g*0.130\frac{J}{g\°C} +60.000g*4.184\frac{J}{g\°C}}\\\\T_{eq}=25.81\°C$

Best regards!

Answer 2

Since the metal sample (lead) was placed in the cup of water, the final temperature of this water is equal to 26.31°C

Given the following data:

• Initial temperature of water = 25.0°C

• Temperature of lead = 67.71°C

• Mass of water = 60.0 grams

• Mass of lead = 37.35 grams

• Specific heat capacity of water = 4.184 J/g°C

• Specific heat capacity of lead = 0.130 J/g°C

To calculate the final temperature of the water:

The quantity of heat lost by lead = The quantity of heat gained by water.

$Q_{lost} = Q_{gained}\\\\mc\theta = mc\theta\\\\37.35(0.130)(67.71) = 60(4.184)(T_f - 25)\\\\328.77 = 251.04(T_f - 25)\\\\328.77 = 251.04T_f - 6276\\\\251.04T_f = 328.77 + 6276\\\\251.04T_f = 6604.77\\\\T_f = \frac{6604.77}{251.04}\\\\T_f = 26.31$

Final temperature of water = 26.31°C

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