An object with a mass of $12 kg$ , temperature of $125 ^oC$ , and a specific heat of $32 (KJ)/(kg*K)$ is dropped into a container with $35 L$ 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 · 2017-05-30T06:48:06

The water does not evaporate and the change in temperature is $=90.5ºC$

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 $DeltaT_o=125-T$

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

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

$C_o=32kJkg^-1K^-1$

$12*32*(125-T)=35*4.186*T$

$125-T=(35*4.186)/(12*32)*T$

$125-T=0.38T$

$1.38T=125$

$T=125/1.38=90.5ºC$

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

An object with a mass of 12 kg12 kg, temperature of 125 ^oC125 ^oC, and a specific heat of 32 (KJ)/(kg*K)32 (KJ)/(kg*K) is dropped into a container with 35 L 35 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?