An object with a mass of $1 kg$ , temperature of $125 ^oC$ , and a specific heat of $32 (KJ)/(kg*K)$ is dropped into a container with $15 L$ 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 · 2017-06-04T06:36:10

The water does not evaporate and the change in temperature is $=42.2ºC$

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 $DeltaT_o=125-T$

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

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

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

$1*32*(125-T)=15*4.186*T$

$125-T=(15*4.186)/(1*32)*T$

$125-T=1.96T$

$2.96T=125$

$T=125/2.96=42.2ºC$

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

An object with a mass of 1 kg1 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 15 L 15 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?