Question

A 0.030-kg ice cube a $0^oC$ is placed in an insulated box that contains a fixed quantity of steam at $100^oC$ . When thermal equilibrium of this closed system is established, its temperature is found to be $23^oC$ . Determine the original mass (in kg) of the steam at $100^oC.$

Answer 1

While the steam condenses to water, the ice melts and then water heats up:

Heat that steam gives out when condenses:

$$Q_s=-m_sq.$$

Heat the steam gives out while cools to 23°C:

$$Q_{ws}=cm_s(23-100).$$

Heat the ice receives when melts:

$$Q_i=m_i L.$$

Heat the water obtained from ice receives while heats:

$$Q_{wi}=cm_i(23-0).$$

The equilibrium of thermal energy looks like this:

$$0=Q_s+Q_i+Q_{w1}+Q_{w2},\\ 0=-m_sq+m_iL+cm_i(23-0)+cm_s(23-100),\\ m_s[q-c(23-100)]=m_i[L+c(23-0)].\\ \\\space\\m_s=m_i\frac{L+c(23-0)}{q-c(23-100)},\\\space\\ m_s=0.03\frac{333550+4200(23-0)}{2264700-4200(23-100)},\\\space\\ m_s=6.64\cdot10^{-3}\text{ kg}.$$

The mass of steam was 6.64 grams.