Question

question 4 (1 point)
100 g of a liquid is 25 °c and is heated to raise the temperature of the liquid to 50.0 °c. the thermal energy gained by the liquid is 6100 j. the specific heat of the liquid is
0.82 j/g °c
2.44 j/g °c
2.06 j/g °c
2.02 j/g °c
2.00 j/g °c

Explanation:

Step1: Write the heat - transfer formula

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature.

Step2: Calculate \(\Delta T\)

\(\Delta T=T_{final}-T_{initial}\). Given \(T_{initial} = 25^{\circ}C\) and \(T_{final}=50^{\circ}C\), then \(\Delta T=50 - 25=25^{\circ}C\). The mass \(m = 100g\) and \(Q = 6100J\).

Step3: Rearrange the formula to solve for \(c\)

From \(Q = mc\Delta T\), we can solve for \(c\) as \(c=\frac{Q}{m\Delta T}\).
Substitute \(Q = 6100J\), \(m = 100g\), and \(\Delta T = 25^{\circ}C\) into the formula: \(c=\frac{6100}{100\times25}\).
Calculate \(\frac{6100}{100\times25}=\frac{6100}{2500}=2.44J/g^{\circ}C\).

Answer:

2.44 J/g \(^{\circ}C\)