An object with a mass of $1 k g$ $1 kg$ , temperature of $105^{o} C$ $105 ^oC$ , and a specific heat of $32 \frac{J}{k g \cdot K}$ $32 J/(kg*K)$ is dropped into a container with $16 L$ $16 L$ of water at $0^{o} C$ $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

Question

1499 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-08T13:50:41

The water does not evaporate and the change in temperature is $= {0.05}^{\circ} C$ $=0.05^@C$

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 $DeltaT_o=105-T$

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

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

The specific heat of the object is $C_o=0.032kJkg^-1K^-1$

The mass of the object is $m_0=1kg$

The mass of the water is $m_w=16kg$

$1*0.032*(105-T)=16*4.186*T$

$105-T=(16*4.186)/(1*0.032)*T$

$105-T=2093T$

$2094T=105$

$T=105/2094=0.05^@C$

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

An object with a mass of 1 kg1kg1 kg, temperature of 105 ^oC105oC105 ^oC, and a specific heat of 32 J/(kg*K)32Jkg⋅K32 J/(kg*K) is dropped into a container with 16 L 16L16 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?