Question

Calculate the qauantity of heat required to raise the temperature of 1g of ice from -10c to 110c

Answer 1

$s_w=1cal/gm°C\\s_{ice}=s_{steam}=\frac{1}{2}Cal/gm°C\\L_w=540cal/gm\\L_{ice}=80cal/gm$

Mass(m)=1g=$10^{-3}kg$

$$Q_1=ms_{ice}∆T=1\times\frac{1}{2}\times10=5cal$$

$Q_2=mL=1\times80=80cal$

$$Q_3=ms_w∆T=1\times1\times100=100cal$$

$Q_4=mL_{w}=1\times540=540cal$

$$Q_5=ms_{steam}∆T=1\times\frac{1}{2}\times10=5cal$$

$Q=Q_1+Q_2+Q_3+Q_4+Q_5$

$$Q=5+80+100+540+5=730cal$$

$Q=730\times4.2=3066J$

Where=1cal=4.2J

$Q=3.066KJ$