Questions: The temperature of a sample of silver increased by 22.4°C when 275 J of heat was applied. What is the mass of the sample? Substance Specific heat J /(g · °C) lead 0.128 silver 0.235 copper 0.385 iron 0.449 aluminum 0.903 m= g

Solution Steps

We are given:

• The temperature increase, \(\Delta T = 22.4^{\circ} \mathrm{C}\)
• The heat applied, \(Q = 275 \, \mathrm{J}\)
• The specific heat of silver, \(c = 0.235 \, \mathrm{J}/(\mathrm{g} \cdot {}^{\circ} \mathrm{C})\)

The formula for heat transfer is: \[ Q = mc\Delta T \] where:

• \(Q\) is the heat added,
• \(m\) is the mass of the sample,
• \(c\) is the specific heat capacity,
• \(\Delta T\) is the change in temperature.

Rearrange the formula to solve for \(m\): \[ m = \frac{Q}{c\Delta T} \]

Substitute \(Q = 275 \, \mathrm{J}\), \(c = 0.235 \, \mathrm{J}/(\mathrm{g} \cdot {}^{\circ} \mathrm{C})\), and \(\Delta T = 22.4^{\circ} \mathrm{C}\): \[ m = \frac{275}{0.235 \times 22.4} \]

Perform the calculation: \[ m = \frac{275}{5.264} \approx 52.24 \, \mathrm{g} \]

Final Answer