Question

q.3 30.0kj of heat is transferred to a 1.00 kg block of aluminum initially at 10.0°c. what will be its final temperature? (ans.43.0°c)
q.4 100g of ethanol at 25°c is heated until it reaches 50°c. how much heat does the ethanol gain? (ans. 6.02x10³j)

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat transferred, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{f}-T_{i}\) ( \(T_{f}\) is the final temperature and \(T_{i}\) is the initial temperature).

For aluminum, \(c = 0.900\space kJ/(kg\cdot^{\circ}C)\), \(m = 1.00\space kg\), \(Q=30.0\space kJ\), and \(T_{i}=10.0^{\circ}C\)

We can rewrite the formula for \(\Delta T\) as \(\Delta T=\frac{Q}{mc}\)

Step2: Calculate \(\Delta T\)

Substitute the values into the formula: \(\Delta T=\frac{30.0\space kJ}{(1.00\space kg)\times(0.900\space kJ/(kg\cdot^{\circ}C))}\)
\(\Delta T=\frac{30.0}{0.900}\space^{\circ}C = 33.3^{\circ}C\)

Step3: Calculate \(T_{f}\)

Since \(\Delta T=T_{f}-T_{i}\), then \(T_{f}=\Delta T + T_{i}\)
\(T_{f}=33.3^{\circ}C+ 10.0^{\circ}C=43.3^{\circ}C\approx43.0^{\circ}C\) (rounding to one decimal place as in the answer)

Answer:

\(43.0^{\circ}C\)