1 Answer

Burgi · Answer 1 · 2023-08-23T02:58:31+0000

Given:

The heat energy added to the sample is Q = 130 J

The mass of each sample is m = 1 kg

The specific heat of aluminum is

$c_A=\text{ 900 J/ kg }^(\circ)C$

The specific heat of brass is

$c_B\text{ = 380 J/kg }^(\circ)C$

The specific heat of glass is

$c_G=\text{ 840 J/kg }^(\circ)C$

The specific heat of steel is

$c_s=450\text{ J/kg }^(\circ)C$

Required: The sample that would require the greatest temperature increase.

Step-by-step explanation:

The temperature increase for a sample can be calculated by the formula

$\Delta T\text{ = }(Q)/(mc)$

On substituting the values, the temperature increase of aluminum will be

$\begin{gathered} \Delta T_A=\text{ }(130)/(1*900) \\ =\text{ 0.144 }^(\circ)C \end{gathered}$

On substituting the values, the temperature increase of brass will be

$\begin{gathered} \Delta T_B=\text{ }(130)/(1*380) \\ =\text{ 0.342 }^(\circ)C \end{gathered}$

On substituting the values, the temperature increase of glass will be

$\begin{gathered} \Delta T_G=\text{ }(130)/(1*840) \\ =\text{ 0.155 }^(\circ)C \end{gathered}$

On substituting the values, the temperature increase of steel will be

$\begin{gathered} \Delta Ts=\text{ }(130)/(1*450) \\ =\text{ 0.289 }^(\circ)C \end{gathered}$

Final Answer: The greatest temperature increase is experienced by brass.

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