Question

during an experiment, scientists place a 4.92 kg cube of an unknown substance into the oven and heat it to test its properties. the cube starts out at room temperature (21.0°c). they measure the heat energy put into the cube to be 14,100 j. when they take the cube out, it is 33.4°c. what is the specific heat capacity of the substance?

231 j/kg°c
174 j/kg°c
412 j/kg°c
330 j/kg°c

Explanation:

Identify the given values

We extract the physical quantities from the problem statement:

• Mass of the cube, \(m = 4.92\text{ kg}\)
• Initial temperature, \(T_i = 21.0^\circ\text{C}\)
• Final temperature, \(T_f = 33.4^\circ\text{C}\)
• Heat energy added, \(Q = 14,100\text{ J}\)

Calculate the temperature change

We find the change in temperature \(\Delta T\) by subtracting the initial temperature from the final temperature:

$$ \Delta T = T_f - T_i $$

$$ \Delta T = 33.4^\circ\text{C} - 21.0^\circ\text{C} = 12.4^\circ\text{C} $$

State the heat transfer formula

The relationship between heat energy, mass, specific heat capacity, and temperature change is given by:

$$ Q = m \cdot c \cdot \Delta T $$

where \(c\) is the specific heat capacity of the substance.

Solve for specific heat capacity

We rearrange the formula to solve for \(c\):

$$ c = \frac{Q}{m \cdot \Delta T} $$

Substitute the known values into the equation:

$$ c = \frac{14,100\text{ J}}{4.92\text{ kg} \cdot 12.4^\circ\text{C}} $$

$$ c = \frac{14,100}{61.008}\text{ J/(kg}\cdot^\circ\text{C)} $$

$$ c \approx 231.12\text{ J/(kg}\cdot^\circ\text{C)} $$

Match with the options

The calculated value of approximately \(231\text{ J/(kg}\cdot^\circ\text{C)}\) matches the first option.

Answer:

• (A) 231 J/kg°C (Correct answer)
• (B) 174 J/kg°C
• (C) 412 J/kg°C
• (D) 330 J/kg°C