An object with a mass of $16 k g$ $16 kg$ , temperature of $270^{o} C$ $270 ^oC$ , and a specific heat of $5 \frac{J}{k g \cdot K}$ $5 J/(kg*K)$ 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

1391 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-29T06:27:49

The water does not evaporate and the change in temperature is $= {0.26}^{\circ} C$ $=0.26^@C$

The heat is transferred from the hot object to the cold water.

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

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

For the object $DeltaT_o=270-T$

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

The specific heat of water is $C_w=4.186kJkg^-1K^-1$

The specific heat of the object is $C_o=0.005kJkg^-1K^-1$

The mass of the object is $m_0=16kg$

The mass of the water is $m_w=32kg$

$16*0.005*(270-T)=32*4.186*T$

$270-T=(32*4.186)/(16*0.005)*T$

$270-T=1674.4T$

$1675.4T=270$

$T=270/1675.4=0.16^@C$

As $T<100^@C$ , the water will not evaporate

An object with a mass of 16 kg16kg16 kg, temperature of 270 ^oC270oC270 ^oC, and a specific heat of 5 J/(kg*K)5Jkg⋅K5 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?