Question

assume the specific heat capacity of water is 4186 j/kg°c.

1. consider mixing 0.15 kg of 30°c water with 0.055 kg of 50°c soya bean oil. the final temperature of the mixture is

(a) 32.94°c
(b) 45.9°c
(c) 58.9°c
(d) 40.0°c
(e) none of the above

Explanation:

Step1: Recall the heat transfer formula

The heat lost by the soya - bean oil is equal to the heat gained by water. The formula for heat transfer is \(Q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Let the final temperature be \(T\). The specific heat capacity of water \(c_{w}=4186\ J/kg^{\circ}C\), and the specific heat capacity of soya - bean oil \(c_{s}=1970\ J/kg^{\circ}C\) (a known value).
The heat gained by water \(Q_{w}=m_{w}c_{w}(T - T_{w})\), and the heat lost by soya - bean oil \(Q_{s}=m_{s}c_{s}(T_{s}-T)\)

Since \(Q_{w}=Q_{s}\), we have \(m_{w}c_{w}(T - T_{w})=m_{s}c_{s}(T_{s}-T)\)

Step2: Substitute the given values

Given \(m_{w} = 0.15\ kg\), \(T_{w}=30^{\circ}C\), \(m_{s}=0.055\ kg\), \(T_{s}=50^{\circ}C\), \(c_{w}=4186\ J/kg^{\circ}C\), \(c_{s}=1970\ J/kg^{\circ}C\)

\(0.15\times4186\times(T - 30)=0.055\times1970\times(50 - T)\)

First, calculate the left - hand side: \(0.15\times4186\times(T - 30)=627.9\times(T - 30)=627.9T-18837\)

Second, calculate the right - hand side: \(0.055\times1970\times(50 - T)=108.35\times(50 - T)=5417.5-108.35T\)

Then, \(627.9T-18837 = 5417.5-108.35T\)

Step3: Solve for \(T\)

Add \(108.35T\) to both sides: \(627.9T + 108.35T-18837=5417.5\)

\(736.25T-18837 = 5417.5\)

Add \(18837\) to both sides: \(736.25T=5417.5 + 18837\)

\(736.25T=24254.5\)

Divide both sides by \(736.25\): \(T=\frac{24254.5}{736.25}\approx32.94^{\circ}C\)

Answer:

A. \(32.94^{\circ}C\)