An object with a mass of $2 kg$ , temperature of $150 ^oC$ , and a specific heat of $24 J/(kg*K)$ is dropped into a container with $18 L$ 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 · 2017-03-04T07:35:22

The water does not evaporate and the change in temperature is $0.01ºC$

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 $DeltaT_o=150-T$

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

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

$C_o=24Jkg^-1K^-1$

$m_0 C_o*(150-T) = m_w* 4186 *T$

$2*24*(150-T)=18*4186*T$

$150-T=(18*4186)/(48)*T$

$150-T=1569.75T$

$1570.75T=150$

$T=150/1570.75=0.01ºC$

The water does not evaporate

An object with a mass of 2 kg2 kg, temperature of 150 ^oC150 ^oC, and a specific heat of 24 J/(kg*K)24 J/(kg*K) is dropped into a container with 18 L 18 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?