An object with a mass of $2 k g$ $2 kg$ , temperature of $311^{o} C$ $311 ^oC$ , and a specific heat of $18 \frac{K J}{k g \cdot K}$ $18 (KJ)/(kg*K)$ is dropped into a container with $37 L$ $37 L$ 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 · 2017-09-16T05:27:41

The water does not evaporate and the change in temperature is $= {58.7}^{\circ} C$ $=58.7^@C$

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 $DeltaT_o=311-T$

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

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

The specific heat of the object is $C_o=18kJkg^-1K^-1$

The mass of the objest is $m_0=2kg$

The mass of the water is $m_w=37kg$

$2*18*(311-T)=37*4.186*T$

$311-T=(37*4.186)/(2*18)*T$

$311-T=4.30T$

$5.30T=311$

$T=311/5.30=58.7^@C$

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

An object with a mass of 2 kg2kg2 kg, temperature of 311 ^oC311oC311 ^oC, and a specific heat of 18 (KJ)/(kg*K)18KJkg⋅K18 (KJ)/(kg*K) is dropped into a container with 37 L 37L37 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?