Question

a metal block with a mass of 200 g releases 4600 j of heat as it cools from 80°c to 30°c. what is the specific heat capacity of the metal?
46 j/g°c
0.46 j/g°c
0.23 j/g°c
0.92 j/g°c
if 2500 j of heat are added to a 100 g sample of ethanol (specific heat capacity = 2.44 j/g°c) initially at 20°c, what will be the final temperature of the ethanol?
30.2°c
50.2°c
10.2°c
40.2°c

Explanation:

Step1: Recall 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 capacity, and \(\Delta T=T_{final}-T_{initial}\) (for heating) or \(T_{initial}-T_{final}\) (for cooling).

Step2: Solve for the specific heat capacity of the metal

For the metal block: \(m = 200g\), \(Q=- 4600J\) (negative because heat is released), \(\Delta T=80 - 30=50^{\circ}C\).
From \(Q = mc\Delta T\), we can solve for \(c\): \(c=\frac{Q}{m\Delta T}\).
Substitute the values: \(c=\frac{-4600J}{200g\times50^{\circ}C}\).
\(c = 0.46J/g^{\circ}C\)

Step3: Solve for the final temperature of ethanol

For ethanol: \(Q = 2500J\), \(m = 100g\), \(c = 2.44J/g^{\circ}C\), \(T_{initial}=20^{\circ}C\).
From \(Q = mc\Delta T\) and \(\Delta T=T_{final}-T_{initial}\), we have \(T_{final}=T_{initial}+\frac{Q}{mc}\).
Substitute the values: \(T_{final}=20^{\circ}C+\frac{2500J}{100g\times2.44J/g^{\circ}C}\).
\(\frac{2500}{100\times2.44}\approx10.2^{\circ}C\)
\(T_{final}=20 + 10.2=30.2^{\circ}C\)

Answer:

For the metal block: \(0.46J/g^{\circ}C\) (the second option).
For the ethanol: \(30.2^{\circ}C\) (the first option).