An object with a mass of $2 kg$ , temperature of $244 ^oC$ , and a specific heat of $33 (KJ)/(kg*K)$ is dropped into a container with $26 L$ 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 · 2017-04-17T12:34:55

The water does not evaporate and the temperature changes by $=92.11ºC$

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 $DeltaT_o=244-T$

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

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

$C_o=33kJkg^-1K^-1$

$2*33*(244-T)=26*4.186*T$

$244-T=(26*4.186)/(2*33)*T$

$244-T=1.649T$

$2.649T=244$

$T=244/2.649=92.11ºC$

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

An object with a mass of 2 kg2 kg, temperature of 244 ^oC244 ^oC, and a specific heat of 33 (KJ)/(kg*K)33 (KJ)/(kg*K) is dropped into a container with 26 L 26 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?