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Answer 1 · 2014-03-30T01:59:29

The final temperature of the water is 49.3 °C.

The formula is #q = mcΔT#, where #q# is the heat, #m# is the mass, #c# is the specific heat capacity, and #ΔT# is the temperature change.

#q# = 455 J; #m# = 32.0 g; #c# = 4.184 J•°C⁻¹g⁻¹; #T_1# = 45.0°C

455 J = #mcΔT# = 25.0 g × 4.184 J•°C⁻¹g⁻¹ × #ΔT#

455 = #104.6ΔT" °C⁻¹"#

#ΔT= 455/("104.6 °C⁻¹")# = 4.350 °C

#ΔT = T_2 – T_1#

#T_2 = T_1 + ΔT# = 45.0 °C + 4.350 °C = 49.3 °C