Question

part a
the specific heat capacity of substance a is twice that of substance b. if samples of equal mass of the two substances absorb the same amount of heat. which substance undergoes the larger change in temperature?
substance a
both substances undergo the same change in temperature
substance b

Explanation:

Step1: Recall the heat - capacity formula

The heat - capacity formula is $Q = mc\Delta T$, where $Q$ is the heat added or removed, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. We can re - arrange it for $\Delta T$: $\Delta T=\frac{Q}{mc}$.

Step2: Analyze the given situation

We are given that $m_A=m_B$ (equal mass), $Q_A = Q_B$ (same amount of heat), and $c_A = 2c_B$ (specific heat capacity of substance A is twice that of substance B).
For substance A, $\Delta T_A=\frac{Q_A}{m_Ac_A}$. For substance B, $\Delta T_B=\frac{Q_B}{m_Bc_B}$.
Substituting $Q_A = Q_B$ and $m_A=m_B$ and $c_A = 2c_B$ into the formulas, we get $\Delta T_A=\frac{Q}{mc_A}$ and $\Delta T_B=\frac{Q}{mc_B}$. Since $c_A = 2c_B$, then $\Delta T_A=\frac{Q}{m\times2c_B}$ and $\Delta T_B=\frac{Q}{mc_B}$. So, $\Delta T_A=\frac{1}{2}\Delta T_B$.

Answer:

C. Substance B