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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