How much heat (in kJ) is required to warm 10.0 g of ice, initially at −10.0 °C, to steam at 110.0 °C? The heat capacity of ice is 2.09 J/g · °C, and that of steam is 2.01 J/g °C.

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princessesther2011 princessesther2011 · Answer 1 · 2020-02-13T06:49:00+01:00

Answer:

4.719 kJ

Explanation:

Q = m(L1∆T + L2T2)

m = 10 g

L1 = 2.09 J/g.°C

L2 = 2.01 J/g.°C

T2 = 110°C

T1 = -10°C

∆T = T2 - T1 = 110-(-10) = 120°C

Q = 10(2.09×120 + 2.01×110) = 10(250.8 + 221.1) = 10(471.9) = 4719 J = 4719/1000 = 4.719 kJ

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snehashish65 snehashish65 · Answer 2 · 2021-11-19T12:59:49+01:00

The amount of heat required to warm the ice is 4.719 kJ.

Given data:

The mass of ice is, m = 10.0 g.

The initial temperature is, T = - 10 degree Celsius.

The final temperature is, T' = 110.0 degree Celsius.

The heat capacity of ice is, c = 2.09 J/g C.

The heat capacity of steam is, c' = 2.01 J/g C.

When ice absorbs heat, the conversion takes place from ice to water (liquid) and from the liquid to steam at 110 degree Celsius. So,

The amount of heat absorbed is given as,

Q = m[c(T' - T) + c'T']

[tex]Q = 10[2.09 (110-(-10)+(2.01 \times 110)]\\\\Q = 4719 \;\rm J[/tex]

Convert in kJ as,

Q = 4719/1000

Q = 4.719 kJ

Thus, we can conclude that the amount of heat required to warm the ice is 4.719 kJ.

Learn more about the thermal equilibrium here:

https://brainly.com/question/13371416