We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

$m=183g$

$l_f=80\frac{cal}{g} =334\frac{J}{g}$

$l_w=1\frac{cal}{g} =4.184\frac{J}{g}$

<em>i) </em>The fusion heat will be:

$Q_f=l_fm=14640cal=14.640kcal$

<em>ii)</em> The heat needed to warm the water from $T_i=0^{\circ}C$ to $T_i=37^{\circ}C$ will be:

$Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal$

So, the total amount needed will be the sum of these two results:

$Q_T=Q_f+Q_w=14.640kcal+6.771kcal=21.411kcal$ .

8 0

2 years ago

Discuss the relationship between amperage, voltage, and power.

Stells [14]

Electrical power, in watts = (voltage, in volts) x (current, in Amperes)

5 0

3 years ago

What is the mass of a 3,500-N rock?

Viktor [21]

350kg because to get Newton’s it’s mass x Gravity, earths gravity is x10 so 3500 divided by 10 is 350

6 0

3 years ago