Question

1. to what temperature will 6500 j of heat raise 7.7 kg of milk that is initially at 25°c. take the specific heat capacity of milk to be 3930 j/kg°c.

(a) 25.9 °c
(b) 20.64°c
(c) 0.64°c
(d) 56.8°c
(e) none of the above

1. cold water at a temperature of 20°c enters a heater, and the resulting hot water has a temperature of 70°c. a person uses 7.9 kg of hot water in taking a shower. assume the specific heat capacity of water is 4186 j/kg°c. the amount of energy needed to heat the water

(a) 2.093×10^6 j
(b) 2.8×10^4 j
(c) 7.7×10^7 j
(d) 1.653×10^6 j
(f) none of the above

Explanation:

Problem 1

Step1: Use heat formula

The formula for heat is \(Q = mc\Delta T\), where \(Q\) is heat, \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T=T - T_0\) ( \(T\) is final temperature, \(T_0\) is initial temperature). We can rewrite it as \(T=T_0+\frac{Q}{mc}\).
Given \(Q = 6500\space J\), \(m = 7.7\space kg\), \(c=3930\space J/kg^{\circ}C\), \(T_0 = 25^{\circ}C\).

Step2: Calculate \(\frac{Q}{mc}\)

\(\frac{Q}{mc}=\frac{6500}{7.7\times3930}\)

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Step3: Calculate \(T\)

\(T=T_0+\frac{Q}{mc}=25 + 0.215=25.215^{\circ}C\)

Step1: Use heat formula

The formula for heat is \(Q = mc\Delta T\), where \(\Delta T=T - T_0\). Here \(T = 70^{\circ}C\), \(T_0=20^{\circ}C\), \(m = 7.9\space kg\), \(c = 4186\space J/kg^{\circ}C\)
\(Q=mc(T - T_0)\)

Step2: Substitute values

\(Q=7.9\times4186\times(70 - 20)\)

$$
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Answer:

(e) none of the above

Problem 2