An object with a mass of $3 k g$ $3 kg$ , temperature of $155^{o} C$ $155 ^oC$ , and a specific heat of $17 \frac{K J}{k g \cdot K}$ $17 (KJ)/(kg*K)$ is dropped into a container with $15 L$ $15 L$ of water at $0^{o} C$ $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-11-30T10:36:31

The water does not evaporate and the change in temperature is $= {69.5}^{\circ} C$ $=69.5^@C$

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 $DeltaT_o=155-T$

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

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

The specific heat of the object is $C_o=17kJkg^-1K^-1$

The mass of the object is $m_0=3kg$

The mass of the water is $m_w=15kg$

$3*17*(155-T)=15*4.186*T$

$155-T=(15*4.186)/(3*17)*T$

$155-T=1.231T$

$2.231T=155$

$T=155/2.231=69.5^@C$

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

An object with a mass of 3 kg3kg3 kg, temperature of 155 ^oC155oC155 ^oC, and a specific heat of 17 (KJ)/(kg*K)17KJkg⋅K17 (KJ)/(kg*K) is dropped into a container with 15 L 15L15 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?