Solution:

We can solve this problem using the concept of heat exchange. Let \( m_h \) be the mass of hot water at 100°C, and \( m_c \) be the mass of cold water at 10°C. The total mass of water is given as 20 kg, so:

\[
m_h + m_c = 20 \, \text{kg}
\]

We also know that the final temperature of the mixture is 35°C. The heat gained by cold water must equal the heat lost by hot water:

\[
\text{Heat gained by cold water} = \text{Heat lost by hot water}
\]

Let the specific heat of water be \( c = 4200 \, \text{J/kg}^\circ \text{C} \). Using the formula for heat change:

\[
m_c c (35 - 10) = m_h c (100 - 35)
\]

Since the specific heat \( c \) is the same for both, it cancels out:

\[
m_c (35 - 10) = m_h (100 - 35)
\]
\[
m_c \times 25 = m_h \times 65
\]

We also know \( m_h + m_c = 20 \), so \( m_c = 20 - m_h \). Substituting this into the equation:

\[
(20 - m_h) \times 25 = m_h \times 65
\]
\[
500 - 25 m_h = 65 m_h
\]
\[
500 = 90 m_h
\]
\[
m_h = \frac{500}{90} \approx 5.56 \, \text{kg}
\]

Thus, the mass of hot water needed is approximately 5.56 kg.