The final temperature of the calorimeter is 34.0 °C.

Two heats are involved:

a. the heat released by the combustion reaction, ##q_1##. b. the heat gained by the calorimeter as it warms up, ##q_2##.

##q_1 + q_2## = 0

##mΔH_”c” + C_”cal”ΔT = 0##

##∆T = T_”f” –T_”i”##, (where ##”f”## and ##”i”## stand for final and initial

##”1.785 g × (-26.42 kJ·g”^-1) + “5.02 kJ°C”^-1 × “(T”_”f” – “24.62) °C = 0″##

##”-47.16 kJ + 5.02 kJ°C”^-1 × “T”_”f” – “123.59 kJ = 0″##

##”5.02 °C”^-1 × T_”f” = 123.59 + 47.16 = 170.75″##

##T_”f” = 170.75/”5.02 °C”^-1 = “34.0 °C”##

The final temperature is 34.0 °C.

Note: The answer can have only three , because that is all you gave for the heat capacity of the calorimeter.. If you need more precision, you will have to recalculate.

The final temperature of the calorimeter is 34.0 °C.

Two heats are involved:

a. the heat released by the combustion reaction, ##q_1##. b. the heat gained by the calorimeter as it warms up, ##q_2##.

##q_1 + q_2## = 0

##mΔH_”c” + C_”cal”ΔT = 0##

##∆T = T_”f” –T_”i”##, (where ##”f”## and ##”i”## stand for final and initial

##”1.785 g × (-26.42 kJ·g”^-1) + “5.02 kJ°C”^-1 × “(T”_”f” – “24.62) °C = 0″##

##”-47.16 kJ + 5.02 kJ°C”^-1 × “T”_”f” – “123.59 kJ = 0″##

##”5.02 °C”^-1 × T_”f” = 123.59 + 47.16 = 170.75″##

##T_”f” = 170.75/”5.02 °C”^-1 = “34.0 °C”##

The final temperature is 34.0 °C.

Note: The answer can have only three , because that is all you gave for the heat capacity of the calorimeter.. If you need more precision, you will have to recalculate.