Given,
first mass m1 = 0.2 kg
temperature t1 = 20°C
Second mass m2 = 0.3 kg
temperature t2 =60°C
final temperature T = ?
specific heat of beaker = S
we know that,
\(T=\;\frac{m_1s_1t_1+m_{2\;}s_2t_2}{m_1s_1+\;m_2s_2}\)
T = (4s +18 s)/ 0.5s
T= 22/0.5
T = 44°C