An object with a mass of $5 kg$ , temperature of $163 ^oC$ , and a specific heat of $14 J/(kg*K)$ 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?

Question

1282 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-27T17:23:13

The water does not evaporate and the temperature of the water changec by $0.09º$ C

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

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

For the object $DeltaT_o=163-T$

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

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

$C_o=14Jkg^-1K^-1$

$m_0 C_o*(163-T) = m_w* 4186 *T$

$5*14*(163-T)=32*4186*T$

$160-T=(32*4186)/(70)*T$

$163-T=1913.6T$

$1914.6T=163$

$T=163/1914.6=0.09ºC$

The water does not evaporate

An object with a mass of 5 kg5 kg, temperature of 163 ^oC163 ^oC, and a specific heat of 14 J/(kg*K)14 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?