Question

what is the final temperature of 25.0 grams of water at 22.0°c after it absorbs 259j of heat? 13.8°c 24.5°c 19.5°c 2.48°c none of the above

Explanation:

Step1: Recall the specific - heat formula

The formula for heat absorption is \(q = mc\Delta T\), where \(q\) is the heat absorbed (\(q = 259J\)), \(m\) is the mass (\(m=25.0g\)), \(c\) is the specific - heat capacity of water (\(c = 4.18J/g^{\circ}C\)), and \(\Delta T=T_f - T_i\) (\(T_i = 22.0^{\circ}C\)).

Step2: Solve for \(\Delta T\)

Rearrange the formula \(\Delta T=\frac{q}{mc}\). Substitute \(q = 259J\), \(m = 25.0g\), and \(c = 4.18J/g^{\circ}C\) into the formula: \(\Delta T=\frac{259J}{25.0g\times4.18J/g^{\circ}C}=\frac{259}{104.5}\approx2.48^{\circ}C\).

Step3: Solve for \(T_f\)

Since \(\Delta T=T_f - T_i\), then \(T_f=\Delta T+T_i\). Substitute \(\Delta T = 2.48^{\circ}C\) and \(T_i = 22.0^{\circ}C\) into the formula: \(T_f=22.0^{\circ}C + 2.48^{\circ}C=24.48^{\circ}C\approx24.5^{\circ}C\).

Answer:

24.5 °C