An object with a mass of $8 kg$ , temperature of $210 ^oC$ , and a specific heat of $24 J/(kg*K)$ is dropped into a container with $36 L$ 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 · 2017-05-27T08:36:28

The water does not evaporate and the change in temperature is $=0.27ºC$

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 $DeltaT_o=210-T$

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

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

$C_o=0.024kJkg^-1K^-1$

$8*0.024*(210-T)=36*4.186*T$

$210-T=(36*4.186)/(8*0.024)*T$

$210-T=784.9T$

$785.9T=210$

$T=210/785.9=0.27ºC$

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

An object with a mass of 8 kg8 kg, temperature of 210 ^oC210 ^oC, and a specific heat of 24 J/(kg*K)24 J/(kg*K) is dropped into a container with 36 L 36 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?