Question

question 14 (5 points)
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a piece of unknown metal weighs 500 g. when the piece of metal absorbs 6.64 kj of heat, its temperature increases from 25°c to 78°c. whats the specific heat capacity of this metal?
a) 25 j/g°c
b) 2.5 j/g°c
c) 0.25 j/g°c
d) 13 j/g°c

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. We need to solve for \( c \), so rearrange the formula to \( c=\frac{q}{m\Delta T} \).

Step2: Convert heat to joules

Given \( q = 6.64\space kJ \), convert to joules: \( 6.64\space kJ\times1000 = 6640\space J \).

Step3: Calculate temperature change

\( \Delta T=T_2 - T_1=78^\circ C - 25^\circ C = 53^\circ C \).

Step4: Substitute values into formula

Mass \( m = 500\space g \), \( q = 6640\space J \), \( \Delta T = 53^\circ C \).
\( c=\frac{6640\space J}{500\space g\times53^\circ C}=\frac{6640}{26500}\approx0.25\space J/g^\circ C \).

Answer:

C) 0.25 J/g°C