Question

1. if no phase change occurs, what is the result of heat flowing into an object?

a. the objects temperature rises
b. the object cools down
c. the objects mass increases
d. the objects density increases

1. an ice cube weighing 0.20 kg initially at 0.0°c is completely melted, and the resulting water is then heated to 53.0°c. how much heat in total is needed to achieve this? ( l_p = 334,000 j/kg, l_v = 2.256\times10^6 j/kg, c = 4.186\times10^3 j/kgcdot^{circ}c ).

a. 111 kj
b. 44 kj
c. 77 kj
d. 11000 kj

1. which of the following statements is true regarding specific heat?

a. it is the same for all materials
b. it only applies to gases
c. it varies depending on the material
d. it is irrelevant to temperature change

Explanation:

1.

Step1: Analyze heat - temperature relationship

When heat flows into an object and no phase change occurs, according to the formula \(Q = mc\Delta T\) (\(Q\) is heat, \(m\) is mass, \(c\) is specific heat capacity, \(\Delta T=T_{final}-T_{initial}\)), if \(Q> 0\) (heat flowing in) and no phase change (so \(m\) and \(c\) are non - zero and constant for the object), then \(\Delta T>0\). This means the object's temperature rises.

Step1: Calculate heat for melting

The heat required to melt the ice is \(Q_1 = m\times L_f\). Given \(m = 0.20\space kg\) and \(L_f=334000\space J/kg\), then \(Q_1=0.20\times334000 = 66800\space J\)

Step2: Calculate heat for heating water

The heat required to heat the water is \(Q_2=mc\Delta T\). Here \(m = 0.20\space kg\), \(c = 4186\space J/(kg\cdot^{\circ}C)\), \(\Delta T=53.0 - 0=53.0^{\circ}C\). So \(Q_2=0.20\times4186\times53.0=44371.6\space J\)

Step3: Calculate total heat

The total heat \(Q = Q_1+Q_2\). \(Q=66800 + 44371.6=111171.6\space J\approx111\space kJ\)

Step1: Recall definition of specific heat

Specific heat \(c\) is defined as \(Q = mc\Delta T\). Different materials have different abilities to absorb or release heat for a given temperature change. For example, water has a high specific heat (\(c = 4186\space J/(kg\cdot^{\circ}C)\)) and metals (like copper, \(c = 385\space J/(kg\cdot^{\circ}C)\)) have lower specific heats.

Answer:

A. The object's temperature rises

2.