= Solution
Use absolute <temperatures> $T_c=278.15\,\mathrm K$ and $T_h=298.15\,\mathrm K$. Let $\dot Q_c$ be the <heat> extracted from the ground, $\dot Q_h$ the <heat> delivered indoors, and $P$ the electrical input <power>. For a reversible <heat pump>, the <entropy> transferred per unit time balances:
$$
\frac{\dot Q_h}{T_h}=\frac{\dot Q_c}{T_c},
\qquad \dot Q_h=\dot Q_c+P.
$$
Consequently the <coefficient of performance> is
$$
\mathrm{COP}_{\rm heating}=\frac{\dot Q_h}{P}
=\frac{T_h}{T_h-T_c}=\frac{298.15}{20}=14.9075.
$$
The requested <heat> delivery is a rate, since the duration is unspecified:
$$
\boxed{\dot Q_h\simeq14.9\,\mathrm{kW}.}
$$
Of this, $13.9\,\mathrm{kW}$ comes from the ground and $1.0\,\mathrm{kW}$ from the electricity. A <coefficient of performance> greater than one is consistent with <conservation of energy> because the <heat pump> transports <heat> as well as converting electrical <energy>.
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