An object with a mass of $6 k g$ $6 kg$ , temperature of $173^{o} C$ $173 ^oC$ , and a specific heat of $14 \frac{J}{k g \cdot K}$ $14 J/(kg*K)$ is dropped into a container with $32 L$ $32 L$ of water at $0^{o} C$ $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

Question

1359 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-05T08:28:35

The water does not evaporate and the change in temperature is $=0.11ªC$

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 $DeltaT_o=173-T$

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

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

$C_o=0.014kJkg^-1K^-1$

$6*0.014*(173-T)=32*4.186*T$

$173-T=(32*4.186)/(6*0.014)*T$

$173-T=1594.7T$

$1595.7T=173$

$T=173/1595.7=0.11ºC$

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

An object with a mass of 6 kg6kg6 kg, temperature of 173 ^oC173oC173 ^oC, and a specific heat of 14 J/(kg*K)14Jkg⋅K14 J/(kg*K) is dropped into a container with 32 L 32L32 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?