300 cal. of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227° C, then the temperature of sink will be 0 C.

Question

300 cal. of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227° C, then the temperature of sink will be 0 C. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

1-( Q 2/ Q 1)=1-( T 2/ T 1) ( Q 2/ Q 1)=( T 2/ T 1) (225/300)=( T 2/500) (500 × 225/300)= T 2 375= T 2 102° C = T 2

Q. $300 c a l$ . of heat is given to a heat engine and it rejects $225 c a l$ . of heat. If source temperature is $22 7^{\circ} C$ , then the temperature of sink will be ______ $^{0} C$ .

$1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{T _{2}}{T _{1}}$
$\frac{Q _{2}}{Q _{1}} = \frac{T _{2}}{T _{1}}$
$\frac{225}{300} = \frac{T _{2}}{500}$
$\frac{500 \times 225}{300} = T_{2}$
$375 = T_{2}$
$10 2^{\circ} C = T_{2}$

Q. 300cal. of heat is given to a heat engine and it rejects 225cal. of heat. If source temperature is 227∘C, then the temperature of sink will be ______0C.

1−Q1​Q2​​=1−T1​T2​​ Q1​Q2​​=T1​T2​​ 300225​=500T2​​ 300500×225​=T2​ 375=T2​ 102∘C=T2​

Q. $300 c a l$ . of heat is given to a heat engine and it rejects $225 c a l$ . of heat. If source temperature is $22 7^{\circ} C$ , then the temperature of sink will be ______ $^{0} C$ .

$1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{T _{2}}{T _{1}}$
$\frac{Q _{2}}{Q _{1}} = \frac{T _{2}}{T _{1}}$
$\frac{225}{300} = \frac{T _{2}}{500}$
$\frac{500 \times 225}{300} = T_{2}$
$375 = T_{2}$
$10 2^{\circ} C = T_{2}$