What is the copper mass ?

Question

A Block of copper has an initial temperature of 65.4C it’s immersed in a beaker containing 95.7 g of H2O. At 22.7C .Once the heat transfer is complete the temperature and block measures 24.2 C. What’s the mass of the copper block ?

Answer 1

Conservation of energy tells us that the heat lost by the hot copper must equal the heat gained by the cooler water. The equation for heat is q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature

heat lost by copper: (using C = 0.385 J/gº obtained from Wikipedia). Also, final temperature of water and Cu will be 24.2º.

q = (m)(0.385 J/gº)(65.4º - 24.2º) = 15.86 x m

heat gained by water: (using 4.184 J/gº

q = (95.7 g)(4.187 J/gº)(24.2º - 22.7º) = 600.6

Setting the two equal and solving for m (mass of copper), we have ...

15.86 x m = 600.6

m = 37.9 g

Answer 2

To solve this problem, you can use the principle of conservation of energy, specifically the heat gained or lost by the water must be equal to the heat gained or lost by the copper block.

The heat gained or lost (�Q) can be calculated using the formula:

�=��Δ�Q=mcΔT

where:

1. �m is the mass,
2. �c is the specific heat capacity,
3. Δ�ΔT is the change in temperature.

For the water: �water=�water⋅�water⋅Δ�waterQwater​=mwater​⋅cwater​⋅ΔTwater​

For the copper block: �copper=�copper⋅�copper⋅Δ�copperQcopper​=mcopper​⋅ccopper​⋅ΔTcopper​

Since the system is isolated, the heat lost by the copper is equal to the heat gained by the water:

�water=�copperQwater​=Qcopper​

Now, you can set up the equation:

�water⋅�water⋅Δ�water=�copper⋅�copper⋅Δ�coppermwater​⋅cwater​⋅ΔTwater​=mcopper​⋅ccopper​⋅ΔTcopper​

Rearrange the equation to solve for �coppermcopper​:

�copper=�water⋅�water⋅Δ�water�copper⋅Δ�coppermcopper​=ccopper​⋅ΔTcopper​mwater​⋅cwater​⋅ΔTwater​​

Now, plug in the given values:

�copper=(95.7 g)⋅(4.18 J/g⋅°C)⋅(24.2 °C−22.7 °C)(0.387 J/g⋅°C)⋅(65.4 °C−22.7 °C)mcopper​=(0.387J/g⋅°C)⋅(65.4°C−22.7°C)(95.7g)⋅(4.18J/g⋅°C)⋅(24.2°C−22.7°C)​

Make sure to convert the mass of water from grams to kilograms (1 g = 0.001 kg) before performing the calculation.