An object with a mass of $18 kg$ , temperature of $270 ^oC$ , and a specific heat of $8 J/(kg*K)$ is dropped into a container with $48 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2018-01-01T08:08:50

THe water will not evaporate and the change in temperature is $=0.19^@C$

The heat is transferred from the hot object to the cold water.

Let $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=T-0=T$

For the object $DeltaT_o=270-T$

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

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

The specific heat of the object is $C_o=0.008kJkg^-1K^-1$

The mass of the object is $m_0=18kg$

The mass of the water is $m_w=48kg$

$18*0.008*(270-T)=48*4.186*T$

$270-T=(48*4.186)/(18*0.008)*T$

$270-T=1395.3T$

$1396.3T=270$

$T=270/1396.3=0.19^@C$

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

An object with a mass of 18 kg18 kg, temperature of 270 ^oC270 ^oC, and a specific heat of 8 J/(kg*K)8 J/(kg*K) is dropped into a container with 48 L 48 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?