An object with a mass of $6 k g$ $6 kg$ , temperature of $50^{o} C$ $50 ^oC$ , and a specific heat of $19 \frac{J}{k g \cdot K}$ $19 J/(kg*K)$ is dropped into a container with $24 L$ $24 L$ 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 · 2017-08-09T14:40:34

The water does not evaporate and the change in temperature is $= {0.06}^{\circ} C$ $=0.06^@C$

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 $DeltaT_o=50-T$

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

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

$C_o=0.019kJkg^-1K^-1$

$6*0.019*(50-T)=24*4.186*T$

$50-T=(24*4.186)/(6*0.019)*T$

$50-T=881.3T$

$882.3T=50$

$T=50/882.3=0.06^@C$

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

An object with a mass of 6 kg6kg6 kg, temperature of 50 ^oC50oC50 ^oC, and a specific heat of 19 J/(kg*K)19Jkg⋅K19 J/(kg*K) is dropped into a container with 24 L 24L24 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?