Question

Steam at 100°C is condensed into a 38.0 g aluminum calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of steam (in g) needed for the system to reach a final temperature of 40.0°C. The specific heat of aluminum is 900 J/(kg · °C).

Answer 1

Answer:

7.2g

Explanation:

From the expression of latent heat of steam, we have

Heat supplied by steam = Heat gain water + Heat gain by calorimeter

mathematically,

$m_{s}c_{w} \alpha _{w}$ + $m_{s}l$=$m_{w}c_{w} \alpha _{w}$ +$m_{c}c_{c} \alpha c_{c}$

L=specific latent heat of water(steam)=2268J/g

$c_{w}$=specific heat capacity=4.2J/gK

$c_{c}$=specific heat capacity of calorimeter =0.9J/gk

$m_{w}$=280g

$m_{c}$=38g

α=change in temperature

$\alpha _{c}$=(40-25)=15

$\alpha _{w}$=(40-25)=15

$\alpha _{s}$=(100-40)=60

Note: the temperature of the calorimeter is the temperature of it content.

From the equation, we can make $m_{s}$ the subject of formula

$m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}$

Hence

$m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g$

Hence the amount of steam needed is 7.2g