An object with a mass of $4 kg$ , temperature of $360 ^oC$ , and a specific heat of $6 J/(kg*K)$ is dropped into a container with $32 L$ 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 · 2017-03-04T15:53:41

The water does not evaporate and the final temperature is $=0.06ºC$

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 $DeltaT_o=360-T$

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

$C_w=4186Jkg^-1K^-1$

$C_o=6Jkg^-1K^-1$

$m_0 C_o*(360-T) = m_w* 4186 *T$

$4*6*(360-T)=32*4186*T$

$360-T=(32*4186)/(24)*T$

$360-T=5581.3T$

$5582.3T=360$

$T=360/5582.3=0.06ºC$

The water does not evaporate

An object with a mass of 4 kg4 kg, temperature of 360 ^oC360 ^oC, and a specific heat of 6 J/(kg*K)6 J/(kg*K) is dropped into a container with 32 L 32 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?