We must identify the heat transfers that are occurring here.
One is the heat transferred from the hot water as it cools (#q_1#).
Another is the heat transferred to the water in the calorimeter as it warms (#q_2#).
The third is the heat transferred to the calorimeter as it warms (#q_3#).
Per the Law of Conservation of Energy, the sum of the three heat transfers must be zero.
#q_1 + q_2 + q_3 = 0#
The formula for the heat absorbed by or released from a substance is
#color(blue)(bar(ul(|color(white)(a/a)q = mCΔTcolor(white)(a/a)|)))" "#
where
#q# is the quantity of heat
#m# is the mass of the substance
#c# is the specific heat capacity of the material
#ΔT# is the temperature change
This gives us
#m_1C_1ΔT_1 + m_2C_2ΔT_2 + C_text(cal)ΔT_3 = 0#
We have
#m_1color(white)(ll) = "49.09 g"#
#C_1 color(white)(ll)= "4.184 J·°C"^"-1""g"^"-1"#
#ΔT_1 = T_"f" - T_"i" = "(42.1 - 72.3) °C" = "-30.2 °C"#
#m_2color(white)(ll) = ?#
#C_2 color(white)(ll)= "4.184 J·°C"^"-1""g"^"-1"#
#ΔT_2 = T_"f" - T_"i" = "(42.1 - 24.2) °C" = "17.9 °C"#
#C_text(cal) = "26.3 J·°C"^"-1"#
#ΔT_3 = T_"f" - T_"i" = "(42.1 - 24.2) °C" = "17.9 °C"#
#q_1 = 49.09 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × ("-30.2" color(red)(cancel(color(black)("°C")))) = "-6203"color(white)(l) color(white)(l)"J"#
#q_2 = m_2 × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × 17.9 color(red)(cancel(color(black)("°C"))) = 74.89m_2color(white)(l) "J·g"^"-1"#
#q_3 = "26.3 J"·color(red)(cancel(color(black)("°C"^"-1"))) × 17.9 color(red)(cancel(color(black)("°C"))) = "470.7 J"#
#q_1 + q_2 + q_3 = "-6203" color(red)(cancel(color(black)("J"))) + 74.89m_2color(red)(cancel(color(black)("J")))·"g"^"-1" + 470.7 color(red)(cancel(color(black)("J"))) = 0#
#74.89 m_2color(white)(l) "g"^"-1" = "6203 - 470.7" = 5732#
#m_2 = 5732/("74.89 g"^"-1") = "76.5 g"#
