Answer
$0.92Kg$
Work Step by Step
We can find the required mass as follows:
For block
$Q_{block}=m_bC_b(T_b)$......eq(1)
For water
$Q_{water}=m_wC_w(T_w)$.......eq(2)
Adding eq(1) and eq(2), we obtain:
$Q_{water}+Q_{block}=0$
$\implies m_wC_w(T_w)+m_bC_b(T_b)=0$
This simplifies to:
$m_b=\frac{m_wC_w(T-T_w)}{C_b(T_b-T)}$
We plug in the known values to obtain:
$m_b=\frac{(1.1)(4186)(22.5-20.0)}{(390)(54.5-22.5)}$
$m_b=0.92Kg$
