Question

question 7 (1 point)
a machine part consists of 0.10 kg of iron and 0.16 kg of copper. how much heat is added to the gear if the temperature increases by 35 c°?
9.1 × 10^2 j
3.8 × 10^3 j
4.0 × 10^3 j
4.4 × 10^3 j

Explanation:

Step1: Recall the heat formula

The heat energy \( Q \) required to change the temperature of a substance is given by \( Q = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For a composite object (iron + copper), we calculate the heat for each component and sum them.

Specific heat capacities: \( c_{\text{iron}} = 450 \, \text{J/(kg·°C)} \), \( c_{\text{copper}} = 387 \, \text{J/(kg·°C)} \), \( \Delta T = 35 \, \text{°C} \), \( m_{\text{iron}} = 0.10 \, \text{kg} \), \( m_{\text{copper}} = 0.16 \, \text{kg} \).

Step2: Calculate heat for iron

For iron: \( Q_{\text{iron}} = m_{\text{iron}}c_{\text{iron}}\Delta T \)
\( Q_{\text{iron}} = 0.10 \, \text{kg} \times 450 \, \text{J/(kg·°C)} \times 35 \, \text{°C} \)
\( Q_{\text{iron}} = 0.10 \times 450 \times 35 = 1575 \, \text{J} \)

Step3: Calculate heat for copper

For copper: \( Q_{\text{copper}} = m_{\text{copper}}c_{\text{copper}}\Delta T \)
\( Q_{\text{copper}} = 0.16 \, \text{kg} \times 387 \, \text{J/(kg·°C)} \times 35 \, \text{°C} \)
\( Q_{\text{copper}} = 0.16 \times 387 \times 35 = 2167.2 \, \text{J} \)

Step4: Total heat

Total heat \( Q = Q_{\text{iron}} + Q_{\text{copper}} \)
\( Q = 1575 + 2167.2 = 3742.2 \, \text{J} \approx 3.8 \times 10^3 \, \text{J} \)

Answer:

\( 3.8 \times 10^3 \, \text{J} \) (corresponding to the option: \( 3.8 \times 10^3 \, \text{J} \))