An ice cute at 0.0*C was dropped into 30.0 g of water in a
cupat 45.0*C. At the instant that all of the ice was melted,
thetemperature of the water in the cup was 19.5*C. What was themass
of the ice cube?

Respuesta :

Answer: The mass of ice cube is 77.90 grams

Explanation:

When ice is mixed with water, the amount of heat released by ice will be equal to the amount of heat absorbed by water.

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]       ......(1)

where,

[tex]m_1[/tex] = mass of ice = ? g

[tex]m_2[/tex] = mass of water = 30.0 g

[tex]T_{final}[/tex] = final temperature = 19.5°C

[tex]T_1[/tex] = initial temperature of ice = 0.0°C

[tex]T_2[/tex] = initial temperature of water = 45.0°C

[tex]c_1[/tex] = specific heat of ice = 2.108  J/g°C

[tex]c_2[/tex] = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

[tex]m_1\times 2.108\times (19.5-0)=-[30.0\times 4.186\times (19.5-45.0)][/tex]

[tex]m_1=77.90g[/tex]

Hence, the mass of ice cube is 77.90 grams