# Question f16a3

##### 2 Answers
Jan 17, 2018

${53.3}^{\circ} C$

#### Explanation:

$\Delta E = m c \Delta \theta$

change in energy = mass * specific heat capacity * temp. change

$305 J = 71.6 g \cdot 0.128 J / g {/}^{\circ} C \cdot \Delta \theta$

$\Delta \theta = \frac{305}{71.6 \cdot 0.128}$

$= 33.3$ (3s.f.)

change in temperature $= {33.3}^{\circ} C$ (3s.f.)

this is a temperature increase.

final temp. = ${20.0}^{\circ} C + {33.3}^{\circ} C$

$= {53.3}^{\circ} C$

Jan 17, 2018

The temperature will increase from ${T}_{i} = {20.0}^{o} C$ to ${T}_{f} = {53.3}^{o} C$

#### Explanation:

The material is heated; thus, it is expected that the temperature of the material will also increase. This problem can be solved using the standard formula:

$Q = m C p \Delta T$
where:

$Q = \text{is the heat/energy}$
$m = \text{ the mass of the material}$
$C p = \text{ the specific heat capacity}$
$\Delta T = \text{the change in temperature}$

where:

DeltaT=T_f-Ti; "final/initial temperatures"

Now, refer to the problem to identify given data that represent each variable in the formula. Per convention, always attached unit to each measured quantity to correctly label it and ensure that units work out and the desired unit is attained as required; thus,

$G i v e n :$

$C p = \frac{0.128 J}{g {\cdot}^{o} C}$
$Q = 305 J$
$m = 71.6 g$
${T}_{i} = {20.0}^{o} C$

$R e q u i red :$

T_f=ul?^oC#

$S o l u t i o n :$

Now, plug in identified data to the standard formula as shown below. But, make sure to rearrange it first to isolate the unknown variable as identified above.

$Q = m C p \Delta T$

where:

$\Delta T = {T}_{f} - {T}_{i}$; plug in this value to the orig. formula

$Q = m C p \left({T}_{f} - {T}_{i}\right)$; isolate the change in temperature, so that

${T}_{f} - {T}_{i} = \frac{Q}{m C p}$; isolate the unknown ${T}_{f}$

${T}_{f} = \frac{Q}{m C p} + {T}_{i}$; thus

${T}_{f} = \frac{305 \cancel{J}}{\left(71.6 \cancel{g}\right) \left(\frac{0.128 \cancel{J}}{\cancel{g} {\cdot}^{o} C}\right)} + {20.0}^{o} C$

${T}_{f} = {33.3}^{o} C + {20.0}^{o} C$

${T}_{f} = {53.3}^{o} C$