An object with a mass of $1 kg$ , temperature of $160 ^oC$ , and a specific heat of $9 J/(kg*K)$ is dropped into a container with $27 L$ 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 · 2017-04-01T06:34:07

The water does not evaporate and the change in temperature is $=0.013ºC$

The heat is transferred from the hot object to 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 $DeltaT_o=160-T$

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

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

$C_o=0.009kJkg^-1K^-1$

$1*0.009*(160-T)=27*4.186*T$

$160-T=(27*4.186)/(1*0.009)*T$

$160-T=12558T$

$12559T=160$

$T=160/12559=0.013ºC$

As $T<100ºC$ , the water does not evaporate.

An object with a mass of 1 kg1 kg, temperature of 160 ^oC160 ^oC, and a specific heat of 9 J/(kg*K)9 J/(kg*K) is dropped into a container with 27 L 27 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?