An object with a mass of $6 kg$ , temperature of $160 ^oC$ , and a specific heat of $9 J/(kg*K)$ is dropped into a container with $27 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

Question

1392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-27T12:27:28

The water does not evaporate, the final temperature is $=0.1ºC$

Let $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=T-0=T$

For the object $DeltaT_o=160-T$

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

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

$C_o=9Jkg^-1K^-1$

$m_0 C_o*(160-T) = m_w* 4186 *T$

$6*9*(160-T)=27*4186*T$

$160-T=(27*4186)/(54)*T$

$160-T=2093T$

$2094T=160$

$T=160/2094=0.1ºC$

The water does not evaporate

An object with a mass of 6 kg6 kg, temperature of 160 ^oC160 ^oC, and a specific heat of 9 J/(kg*K)9 J/(kg*K) is dropped into a container with 27 L 27 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?