This is a typical calorimetry question.

There are two heat transfers involved:

#"heat lost by Cu + heat gained by water" = 0#

#color(white)(mmmm)q_1 color(white)(mmll)+color(white)(mmmm) q_2color(white)(mmmmm) =0#

#color(white)(mm)m_1C_1ΔT_1 color(white)(m)+color(white)(mm) m_2C_2ΔT_2 color(white)(mmm)= 0#

Let's calculate each heat separately.

#q_1 = m_1C_1ΔT_1 = 250. color(red)(cancel(color(black)("g"))) × "0.385 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × (T_text(f) - "100 °C") = "96.25 J·°C"^"-1"(T_text(f) - "100 °C") = 96.25T_text(f) color(white)(l)"J·°C"^"-1" - "9625 J"#

#q_2 = m_2C_2ΔT_2 = 500 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × (T_text(f) - "15 °C") = "2092 J·°C"^"-1"(T_text(f) - "15 °C") = 2092T_text(f) color(white)(l)"J·°C"^"-1" - "31 380 J"#

Now, we add the two heats and combine like terms.

#q_1 + q_2 = 96.25T_text(f) color(red)(cancel(color(black)("J")))·"°C"^"-1" - "9625" color(red)(cancel(color(black)("J"))) + 2092T_text(f) color(red)(cancel(color(black)("J")))·"°C"^"-1" - "31 380" color(red)(cancel(color(black)("J"))) = 0#

#2188T_text(f)color(white)(l) "°C"^"-1" - "41 005" = 0#

#T_text(f) = "41 005"/("2188 °C"^"-1") = "18.7 °C"#