An object with a mass of $4 kg$ , temperature of $284 ^oC$ , and a specific heat of $12 (KJ)/(kg*K)$ is dropped into a container with $25 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-02-22T12:08:16

Water will not evaporate, the final temperature is $=89.3$ ºC

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

Let $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=T-0=T$

For the object $DeltaT_o=284-T$

$m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)$

$C_w=4.186kJkg^-1K^-1$

$C_o=12kJkg^-1K^-1$

$m_0 C_o*(270-T) = m_w* 4.186 *T$

$4*12*(284-T)=25*4.186*T$

$284-T=(25*4.186)/(48)*T$

$284-T=2.18T$

$3.18T=284$

$T=284/3.18=89.3ºC$

An object with a mass of 4 kg4 kg, temperature of 284 ^oC284 ^oC, and a specific heat of 12 (KJ)/(kg*K)12 (KJ)/(kg*K) is dropped into a container with 25 L 25 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?