Problem with unit coversion (temperature) I'm having some difficulty understanding one formula that I was introduced to today. The problem concerns converting BTU's to joules.
The formula given by my teacher as well as Wikipedia says that:
$1 BTU = 1lb \cdot c \cdot\Delta 1 ^{\circ}F$ ; 
where lb is pound-mass and c is the specific heat capacity. 
For water we know that: $c=4,185 \cfrac{kJ}{kg\cdot\Delta^{\circ}C }$. Now I want to prove that $1$ BTU is equal to $1,056$ kJ. In the following transformation I'm going to be using the fact that $\Delta^{\circ}F = \cfrac 95\Delta^{\circ}C$
$1BTU = 1[lb] \cdot c \cdot\Delta 1 ^{\circ}F = 0,454 [kg]  \cdot 4,185 \cfrac{kJ}{kg\cdot\Delta^{\circ}C }\cdot \Delta1 ^{\circ}F = \\ 0,454 [kg]  \cdot 4,185 \cfrac{kJ}{kg\cdot\cfrac59\Delta^{\circ}F }\cdot \Delta 1 ^{\circ}F$
After simplifying, we get that $1$ BTU is equal to $0,454 \cdot\cfrac95\cdot4,185[kJ]$; The correct anwser is that 
$1$ BTU is equal to  $0,454 \cdot\cfrac59\cdot4,185[kJ]$
I honestly cannot spot the mistake. Any help appreciated!
 A: In your derivation, I don't see where there could be an error. However, I think the problem is clarified by thinking about the specific heat capacity in degrees Fahrenheit.
Specific heat capacity can be seen in the following way: "Informally, it is the amount of energy that must be added, in the form of heat, to one unit of mass of the substance in order to cause an increase of one unit in its temperature." [Wikipedia]
A change in degrees Celsius equals $\frac{5}{9}$ times that change in degrees Fahrenheit (as you said $\Delta ^{\circ} F = \frac{9}{5} \Delta^{\circ}C$). For a $\frac{5}{9}$ unit increase in temperature, $\frac{5}{9}$ of the amount of energy must be added. Therefore, to express the specific heat capacity in Fahrenheit, it is multiplied by $\frac{5}{9}$ (instead of divided). This results in the desired $0,454\cdot \frac{5}{9} \cdot 4,185 [kJ]$.
A: I suspect that the confusion is all about the definition of "specific heat". Let me try to avoid that altogether.
Observe that
$Q \propto m$ and $Q \propto T$ $\implies Q \propto mT$
where $Q = $ heat transfer, $m = $ mass of the substance and $T = $ change in temperature
Equivalently, $\frac{Q}{mT} = $ constant
In other words, $\frac{Q_1}{m_1T_1} = \frac{Q_2}{m_2T_2} = \ldots$
Now consider two cases:
$Q_1 = 1$ Cal $ = 4185$J, $m_1 = 1$kg and $T_1 = 1^\circ$C
and
$Q_2 = x$ J, $m_2 = 0.454$kg and $T_2 = \frac{5}{9}^\circ$C
Next solve
$\frac{x}{0.454 \times \frac{5}{9}} = \frac{4185}{1 \times 1}$
Finally $Q_2 = 1055.55$J
