Question

l_{v}=540cal/g
l_{f}=80cal/g

1. consider mixing 100 g of 25°c water with 75 g of 40°c ethanol. take the specific heat

capacity of ethanol as 2.4 j/g°c and the specific heat capacity of water as 1 j/g°c. the final
temperature of the mixture is
(a) 56.9°c
(b) 43.8°c
(c) 34.8°c
(d) 45.9°c
(e) none of the above

Explanation:

Step1: Set up heat transfer equation

According to the principle of heat transfer \(Q = mc\Delta T\), and the heat lost by ethanol is equal to the heat gained by water. Let the final temperature be \(T\).
The heat lost by ethanol: \(Q_{lost}=m_{ethanol}c_{ethanol}(T_{ethanol}-T)\)
The heat gained by water: \(Q_{gained}=m_{water}c_{water}(T - T_{water})\)
So, \(m_{ethanol}c_{ethanol}(T_{ethanol}-T)=m_{water}c_{water}(T - T_{water})\)

Step2: Substitute values

Given \(m_{ethanol} = 75g\), \(c_{ethanol}=2.4J/g^{\circ}C\), \(T_{ethanol}=40^{\circ}C\), \(m_{water}=100g\), \(c_{water}=1J/g^{\circ}C\), \(T_{water}=25^{\circ}C\)
Substitute into the equation:
\(75\times2.4\times(40 - T)=100\times1\times(T - 25)\)
\(180\times(40 - T)=100\times(T - 25)\)
\(7200-180T = 100T-2500\)
\(7200 + 2500=100T + 180T\)
\(280T=9700\)
\(T=\frac{9700}{280}\approx34.64^{\circ}C\approx34.8^{\circ}C\)

Answer:

C. \(34.8^{\circ}C\)