# Question 41af6

Mar 30, 2014

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