An object with a mass of $6 k g$ $6 kg$ , temperature of $240^{o} C$ $240 ^oC$ , and a specific heat of $5 \frac{J}{k g \cdot K}$ $5 J/(kg*K)$ is dropped into a container with $15 L$ $15L$ 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 · 2017-07-18T13:16:01

The water does not evaporate and the change in temperature is $= {0.11}^{\circ}$ $=0.11^@$

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 $DeltaT_o=240-T$

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

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

$C_o=0.005kJkg^-1K^-1$

$6*0.005*(240-T)=15*4.186*T$

$240-T=(15*4.186)/(6*0.005)*T$

$240-T=2093T$

$2094T=240$

$T=240/2094=0.11^@C$

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

An object with a mass of 6 kg6kg6 kg, temperature of 240 ^oC240oC240 ^oC, and a specific heat of 5 J/(kg*K)5Jkg⋅K5 J/(kg*K) is dropped into a container with 15L 15L15L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?