Answer:
1.2 °C
Step-by-step explanation:
The formula relating the heat (q) absorbed by an object, the mass (m), and the temperature change (ΔT) is
q = mCΔT Divide both sides by mC
ΔT = q/(mC)
Data:
q = 15 J
m = 3.0 g
C = 4.184 J·°C⁻¹g⁻¹
Calculation:
ΔT = 15/(3.0 × 4.184)
ΔT = 15/12.6
ΔT = 1.2 °C
The change in temperature of the sample has been [tex]\rm 1.2\;^\circ C[/tex].
The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.
The expression for specific heat has been given as:
[tex]q=mc\Delta T[/tex]
For the given water sample,
The mass of sample has been [tex]m=3\;\rm g[/tex]
The heat absorbed by the sample has been [tex]q=15\;\rm J[/tex]
The specific heat of water has been [tex]c=4.184 \;\rm J./^\circ C.g[/tex]
Substituting the values for change in temperature, [tex]\Delta T[/tex]:
[tex]15=3\;\times\;4.184 \;\times\;\Delta T\\\Delta T=\dfrac{15}{3\;\times\;4.184 } \\\Delta T=1.2\;^\circ \rm C[/tex]
The change in temperature of the sample has been [tex]\rm 1.2\;^\circ C[/tex].
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