IBDP Physics B.1 Thermal energy transfers IB Style Question Bank : HL Paper 1

Ans:B

The change in thermal energy (\(Q\)) is related to the mass of the material, the specific heat capacity (\(c\)), and the change in temperature (\(\Delta T\)) by the equation:

\[Q = mc\Delta T\]

For both cubes \(X\) and \(Y\), made of the same material, the specific heat capacity is the same (\(c\)).

Now, let’s consider cube \(X\). Its length is \(L\), so its volume (\(V_X\)) is \(L^3\). Cube \(Y\) has twice the length (\(2L\), so its volume (\(V_Y\)) is \((2L)^3 = 8L^3\).

Since the mass (\(m\)) is directly proportional to volume and density, and both cubes are made of the same material, we can say that:

\[m_X = k \cdot V_X\]
\[m_Y = k \cdot V_Y\]

Where \(k\) is a constant.

So, the ratio of the mass of cube \(Y\) to the mass of cube \(X\) is:

\[\frac{m_Y}{m_X} = \frac{k \cdot V_Y}{k \cdot V_X} = \frac{8L^3}{L^3} = 8\]

Now, let’s consider the thermal energy gained:

For cube \(X\): \(Q_X = m_Xc\Delta T\)

For cube \(Y\): \(Q_Y = m_Yc\Delta T_Y\)

Given that \(Q_Y = 2Q_X\) and \(\frac{m_Y}{m_X} = 8\), we can write:

\[2Q_X = 8(m_Xc\Delta T)\]

Now, we can simplify:

\[2Q_X = 8Q_X\]

Divide both sides by 2:

\[Q_X = 4Q_X\]

This implies that \(\Delta T_Y\) for cube \(Y\) must be four times the change in temperature \(\Delta T\) for cube \(X\).

So,

\[\Delta T_Y = 4\Delta T\]

The correct answer is B. \(\frac{\Delta T}{4}\).